UC-NRLF 


Fs 


B  ^  ^^a  ibfl 


menI^I^WPi 


^^T^WSS^^T^^ 


IIV<M 


n 


R  S  T 


I  i 


'HER  Sower  Gompany 


MENSURATION 


WITH  SPECIAL  APPLICATION  OF  THE 
PRISMOIDAL    FORiMULA 


Designed  for  Supplementary  Work  in 

Normal  Schools,  High  Schools,  Academies,  and  Advanced 

Classes  in  Grammar  and  Common    Schools 


BY 

S.    W.   FURST 


PHILADELPHIA 

CHRISTOPHER    SOWER     COMPANY 
1911 


MENSURATION 


WITH  SPECIAL  APPLICATION  OF  THE 
PEIS3I0IDAL   F0E3IULA 


Designed  for  Supplementary  Work  in 


Classes  in  Grammar  and  Common    Schools 


BY 

S.    W.   FURST 


phtladelphta 

CHRISTOPHER    SOWER     COMPANY 

1911 


F 


Copyright,  1898,  1900 
By  S.  W.  FUPuST 


PREFACE 

A  NEW  text-book  at  the  close  of  the  luiieteeuth  cen- 
tury is  no  novelty'.  The  author,  therefore,  without  apol- 
ogy sends  forth  this  little  volume,  hopin<^  that  his  co- 
workers nia}'  find  at  least  some  of  the  pleasure  and 
profit  in  using  it  that  he  has  found  in  its  preparation 
and  practical  application  in  the  school -room. 

The  method  of  treating  Mensuration,  as  here  presented, 
has  had  a  thorough  class-room  test — a  safe  criterion  for 
any  method.  Both  those  pupils  with  a  knowledge  of 
Geometry  and  those  without  such  knowledge  have  been 
able  to  solve  problems  with  facility.  Without  this  test 
with  pupils  of  various  ages  and  grades,  the  author  would 
hesitate  long  before  advising  the  setting  aside  of  some 
of  the  standard  rules  used  in  text -books  for  years. 
Pupils  who  have  become  familiar  wdth  this  method  have 
invariably  become  enthused  witli  the  work.  They  enter 
examinations  in  this  department  confident  of  success, 
because  they  have  become  masters  of  one  broad  prin- 
ciple covering  a  great  field  of  measurement.  What  was 
vague  becomes  clear;  the  concrete  supplants  the  abstract. 
Concentration  and  simplicity  are  everywhere  the  central 
ideas. 

A  few  words  concerning  the  scope  of  this  principle 
may  not  be  amiss.  As  the  title-page  indicates,  the 
special  feature  of  this  w^ork  is  the  application  of  the 
prismoidal  formula  to  finding  the  volume  of  solids.  This 
is  used  by  civil  engineers  in  making  calculations  for 
railway  embankments,  excavations,  etc.;  but,  so  far  as 
we    are    aware,    no    common  s<'bool    text -book    treats    the 


260bb8 


vi  PREFACE 

subject  at  leugth.  It  applies  to  many  other  solids  than 
those  which  fall  strictly  within  the  definition  of  a  pris- 
nioid.  Ellwood  Mivrris,  a  civil  engineer,  made  known 
this  extended  application  about  1840: 

"It  applies  to  all  solids  having  two  parallel  sides  or 
faces,  where  these  two  faces,  curved  or  plane,  are  united 
by  surfaces  upon  which  and  through  every  point  of  which 
a  straight  line  may  be  made  to  connect  the  two  parallel 
faces;  to  prisms,  wedges,  pyramids,  cones,  etc.,  regular 
or  irregular,  right  or  oblique;  to  the  sphere,  hemisphere, 
and  other  spherical  segments;  to  either  section  of  a 
cylinder,  where  the  cutting  plane  is  passed  through  bott 
parallel  sides  ;  to  any  section  of  a  cone,  where  the  cut- 
ting plane  is  passed  through  both  apex  and  base ;  to 
frustums  of  prisms,  pyramids,  cones,  etc."  Volumes  of 
frustums  of  pj-ramids  and  cones  can  -be  found  by  its 
use  without  the  extraction  of  a  root.  It  is  thus  seen 
that  its  scope  is  very  great.  For  the  derivation  of 
this  formula  from  the  principles  of  Geometry,  see 
Supplement. 

Care  has  been  taken  to  use  only  such  rules  and  defi- 
nitions as  have  been  sanctioned  by  the  best  authorities. 
In  order  that  comparisons  may  be  readily  made  without 
reference  to  other  books,  and  that  the  teacher  may  con- 
veniently make  selections  of  those  which  best  suit  his 
purpose,  a  number  of  these  are  given  under  each 
chapter. 

The  work  abounds  in  interesting  problems,  many  of 
which  have  been  suggested  by  objects  familiar  to  every 
boy  and  girl.  Actual  measurements  were  made  for  many 
of  them.  Both  the  grading  of  the  problems  and  their 
practical  utility  have  been  kept  in  mind.  It  is  believed 
that  enough  neiv  ones  have  been  introduced  to  awaken 
interest  in  the  dullest  boy. 


nuci'AcE  vii 

It  is  iuteiult'd  that  this  work  be  riupplemeutal  to  both 
text -books  on  Arithmetic  and  Geometry.  Many  text- 
books on  Geometry  are  so  abstract  in  treatment  as  to 
leave  the  pupil,  even  after  completing  his  course,  unable 
to  apply  what  he  has  learned.  Few  persons  can  easily 
acquire  the  power  of  abstraction  necessar}^  for  correct 
geometrical  conceptions.  Making  the  subject  concrete, 
therefore,  will  be  of  great  benefit  to  the  majority  of 
pupils.  As  soon  as  the  pupil  discovers  the  practical 
value  of  the  ..propositions  he  demonstrates,  he  will  be 
incited  to  study  with  new  interest  and  zeal. 

Fundamentals  of  Measurement,  the  frontispiece,  can  be 
made  of  much  use  to  both  teacher  and  pupil.  It  will  be 
noticed  that  the  fundamental  quantities  of  measurement — 
points,  lines,  surfaces,  and  volumes — are  so  grouped 
as  to  picture  with  great  vividness  the  relations  of  these 
fundamentals.  Each  block  is  independent  of  the  others, 
and  yet  it  requires  all  the  blocks — and  in  the  order 
given — to  show  systematically  the  essentials  of  measure- 
ment. By  starting  at  the  top  of  the  monument  and 
noticing  in  order  the  point,  lines,  plane  surfaces,  and 
solids,  the  synthetic  process  of  reasoning  becomes  ap- 
parent. The  simplicity  of  the  arrangement  is  surpassed 
only  by  the  comprehensiveness  of  the  whole.  By  it  the 
beginner  can  learn  the  elements,  and  the  advanced 
student,    at    a    glance,    verify    his    conclusions. 

The  author  tenders  his  thanks  to  Charles  Lose,  A.M., 
Superintendent  of  Schools  of  the  city  of  Williamsport, 
Pa.,  the  Hon.  Emerson  Collins,  of  tlie  same  place,  and 
others,  who  have  kindly  given  him  valuable  suggestions 
and  criticisms  during  the  progress  of  the  work.  He 
asks  from  his  fellow  teachers  not  only  a  hearty  welcome, 
but  criticisms  as  well. 

S.  W.  FURST. 


SUGGESTIONS   TO   TEACHERS 

It  is  deemed  best  to  show  the  application  of  the 
formula  to  those  solids  for  which  the  rules  usually  given 
are  short  and  simple,  as  well  as  to  those  for  which  the 
rules  given  are  more  complex.  This  is  done  by  means 
of  a  chapter  of  illustrative  problems,  w^hich  can  be  used 
•for  reference  for  all  following  chapters  on  solids.  It  is 
believed  that  a  plan  for  the  solution  of  every  problem  in 
the  book  can  be  found  by  a  study  of  this  chapter.  After 
the  pupil  thoroughly  understands  how  to  apply  the  for- 
mula to  the  solution  of  problems  involving  all  the  com- 
mon solids,  it  is  well  to  have  him  use  in  practice  the 
shorter  rules,  especially  for  finding  the  volume  of  the 
prism,  cylinder,  etc. 

The  pupil  should  be  thoroughly  drilled  on  finding  the 
area  of  the  middle  section.  This  can  best  be  done  by 
liaving  him  draw  figures  of  different  shapes  and  repre- 
senting this  section  by  dotted  lines.  In  all  cases  he 
should  write  on  each  line  the  dimension  given  in  the 
solid.  Until  he  is  able  to  do  this  preliminary  work,  it 
can  not  be  expected  that  he  will  make  rapid  progress. 
The  plan  is  made  clear  in  the  chapter  of  illustrative 
problems. 

Constant  reference  should  be  made  to  the  frontispiece. 
Many  points  often  vague  to  the  pupil  can  be  made  quite 
clear  by  a  proper  use  of  this.  It  is  suggested  that  the 
pupil  be  taught  to  draw  and  letter  this  from  memory, 
a  block  at  a  time,  and  fiuallv  the  whole  monument. 


(viii) 


CONTENTS 


PAGE 

Preface v 

Suggestions  to  Teachers viii 

Introduction 1 

Mensuration  Defined 2 

Lines 3 

Angles   3 

Plane  Figures  Defined 4 

Mensuration  of  Surfaces  — 

The  Triangle 5 

The  Right-Angled  Triangle 8 

Quadrilaterals 11 

The  Circle 14 

The  Ellipse 18 

Similar  Plane  Figures 19 

Mensuration  of  Solids  — 

Prisms  and  Cylinders 22,  24 

Surfaces  of  Prisms  and  Cylinders 25 

Illustrative  Problems  Solved  by  the  Prismoidal  Formula  .  27 

Volumes  of  Prisms  and  Cylinders 37 

Pyramids  and  Cones 39 

Surfaces  of  Pyramids  and  Cones  and  Their  Frustums    .    .  42 

Volumes  of  Pyramids  and  Cones 44 

Volumes  of  Frustums  of  Pyramids  and  Cones 46 

The  Sphere 48 

The  Spheroid 52 

Circular  Rings 55 

The  Wedge 56 

Similar  Solids 58 

Miscellaneous  Problems 60 

Supplement 66 

(ix) 


MENSURATION" 


Mensuration  was  formerly  a  part  of  the  course  of  study 
of  everv  well-ordered  school  of  higher  grade.  The  addition 
of  a  variety  of  subjects  to  the  course  of  study  gradually 
crowded  it  out,  so  that  it  often  became  merely  an  insignifi- 
cant part  of  the  arithmetic  work. 

Now  a  reaction  has  set  in.  Tlie  demand  for  thorough- 
ness instead  of  variety  is  growing,  and  the  great  increase  in 
^^  Industrial  Education  "  has  awakened  an  interest  in  men- 
suration as  a  practical  subject  and  has  called  for  increased 
attention  to  it. 

A  knowledge  of  mensuration  is  valuable  to  : 

The  housekeeper  who  buys  wall-paper,  curtains,  carpets, 
etc. 

The  farmer  who  measures  his  fields  and  bins  to  count  his 
crop  values. 

The  carpenter  who  builds  houses,  barns,  or  shops. 

The  mason  who  builds  walls  and  paves  sidewalks. 

The  plumber  who  calculates  pipe  extension,  water  flow, 
and  drainage. 

The  paiuter  who  paints  buildings  and  fences  and  deco- 
rates interiors. 

The  contractor  who  paves  streets  and  builds  roads. 

The  coo})er  who  makes  kegs,  casks,  and  barrels. 

The  landscape  gardener  who  lays  out  grounds  and  flower 
beds. 

The  average  man  who  employs  artisans. 


MENSURATION 


INTRODUCTION 


1.  A    Brick   (Fig.   1)    is    a    material    bodj",    since    it 
occupies  bounded  space.    The  bound- 
ary    of     the     bricK:    is    its     surface.  Brick 
We   can   conceive    the    brick    to     be 
removed,    and    there     would    remain 
space  formerly  occupied  by  the  brick;  ng.i 
this  is  a  geometrical  solid.     The  sur- 
face   bears    the    same    relation    as    in   the  material    body. 

2.  A  Plane  is  a  surface  such  that  a  straight  line  con- 
necting any  two   of  its  points 

lies  wholly  in  its  surface. 
Thus,  MN  (Fig.  2)  is  a  plane, 
since  all  points  of  any  straight 
line,  as  RS  ,  lie  wholly  in  its 
surface.  Each  face  of  the 
brick  (Fig.  1)   is  a  plane. 


Fig.  2 


3.  When  two  planes  intersect  each  other,  their  com- 
mon intersection  is  a  straight  line. 
Each  edge  of  the  brick  (Fig.  1)  is  a 
straight  line,  since  in  each  case  it  is 
the  intersection  of  two  planes.  RS 
(Fig.  3)  is  a  straight  line,  it  being  the 
common^  intersection  of  the  two  planes 
AB  and  CD. 

A  CI) 


2  ML'ysrh'JTiox 

4.  A  Straight  Line  is  one  which  has  the  same  direc- 
tion at  every  point.     AB  (Fig.  4) 

is     a     straight     line.     When     two  "^  Ji^  ^ 

straight  lines  intersect,  their  com- 
mon intersection  is  a  point.  In 
Fig.  5  R  is  a  point,  it  being  the 

intersection    of    the    two    straight  ^ — -7 \ 

lines  AB  and  CD.  ^      ^'^'  ^ 

5.  A  Point   has  no  extension,  but  simply  position. 
A    Line  has  only  one  dimension, —  length. 

A    Surface  has  two  dimensions, —  length  and  breadth. 

A  Solid,  or  volume,  has  three  dimensions, —  lengthy 
'breadth,,  and   thickness. 

The  brick  (Fig.  1)  is  a  solid. 

Any  face  of  the  brick  is  a  surface. 

Any  edge  of  the  brick  is  a  line. 

Any  corner  of  the  brick  is  a  point. 

A  line  is  limited  by  a  point ;  a  surface  is  limited  by 
a  line  ;  a  volume  is  limited  by  a  surface  ;  and  yet  we 
can  conceive  of  each  of  these  as  being  independent  of 
the  other. 

Points,  lines,  surfaces,  and  solids  of  geometry  are 
purely  imaginary. 

The  fundamental  quantities  of  measurement  are 
points,  lines,  surfaces,  and    volumes. 

6.  Mensuration  is  that  branch  of  applied  geometry 
which  gives  the  rules  and  processes  for  finding  the 
lengtlis  of  lines,  the  areas  of  surfaces,  or  the  volumes 
of   solids. 


DEFINITIONS 


LINES 

7.  A  Straight  Line  is  one  which  has  the  same  direc- 
tion at  all  points  :    as  MN.     It  is  the 

,  M N 

shortest   distance   between  two   points. 

8.  A  Curved  Line  is  one  which 
changes  its  direction  at  every  point  : 
as  RS. 

9.  A  Broken  Line  is  one  which 
is  composed  of  different  successive 
straight  lines  :   as  ABODE. 

10.  Parallel  Lines  are  those  in  the  same  plane  which 
have  the  same  direction  :    as  MN  and 

ST.      They   can    never   meet,  however       p  ^ 

far  produced. 

11.  A    Perpendicular    Line    is    a 

straight  line  joined  to  another  straight 
line,  so  as  to  incline  equally  to  both 
sides.     AB  is  perpendicular  to  CD. 


ANGLES 

12.    An    Angle  is    the  difference  in   direction   between 
two  lines  which  meet    at    a  common    point,  ^a 

called  the  vertex.     ABC   is  an  angle  whose 
vertex    is   the    point    B,    and    w^iose    sides     ^' 
are  AB  and  BC. 

(3) 


MKXsriiATK/X 


13.  A  Rig*ht  Ang-le  is  an  angle 
formed  by  two  lines  which  are  perpen- 
dicular to  each  other.  ABC  is  a  right 
angle. 

14.  An    Acute     Ang-le    is    one 

which     is    less    than    a    right    angle. 
MNO  is  an  acute  angle. 

15.  An  Obtuse  Ang'le  is  one  which  is  greater  than  a 
right  angle.    ABC  is  an  obtuse  angle.        a 

Acute     and     obtuse      angles      are  N. 

oblique  angles.  b  ^ 


PLANE   FIGURES 


16.  A    Plane  is    a  surface  such   that  a   straight  line 
connecting  any  two  of  its  points 
lies  wholly  in  its  surface.     MN  /    vR 
is   a   plane,  since    all   points   of            /R- 
the  straight  line  RS   lie  wholly 
in  its  surface. 

17.  A  Plane  Fig'ure  is  a  portion  of  a  plane  surface 
bounded  by  straight  or  curved  lines. 

18.  A  Polygfon  is  a  plane  figure  bounded  by  straight 
lines;   the  lines  are  called  the  sides  of  the  polygon. 

19.  The   Perimeter  of  a   polygon   is   the   sum   of    its 
sides. 

20.  The    Area   of   a   plane    figure   is   the   surface    in- 
cluded within  the   perimeter. 

21.  The  Diag'Onal  of  a  polygon  is  a  line  joining  the 
vertices  of  any  two  of  its  angles  not  adjacent. 

22.  The  Altitude  of  a  polygon  is  the  perpendicular 
distance  between  its  base  and  a  side  or  angle  opposite. 


TRIAMiLES  5 

23.  An  Equilateral  Polygron  has  all  its  sides  equal. 
An  Equiang-ular  Polyg-on  has  all  its  angles  equal. 

A    Reg^ular    Polyg-on    is    both    equilateral    and   equi- 
angular. 

24.  Polygons  are  named  according  to  the  number  of 
sides.     A  polygon   of   three   sides   is  a  triangle  ;    of  four 


Triangle 


Quadrilateral 


Pentagon 


Hexagou 


Heptagou 


Octagon 


Isonagon 


Decagon 


sides,   a    quadrilateral  ;     of    five    sides,   a    pentagon ;  of 

six    sides,   a   hexagon  ;    of   seven    sides,  a   heptagon  ;  of 

eight   sides,  an  octagon  ;    of   nine   sides,  a  nonagon  ;  of 
ten  sides,  a  decagon;  etc. 


TRIANGLES 

25.  A  Triang'le  is  a  plane  figure  having  three  sides 
and  three  angles.     ABC  is  a  triangle.  ^ 

26.  The  Base  of  a  triangle  is  the  side 
on  which  it  stands.     AB  is  the  base. 


27.    The  Altitude  of  a  triangle  is  a  line 
drawn  from  the  angle  opposite  perpendicular  to  the  base, 


6 


MEysuitATioy 


or  the  base  produced.     CD    is  the  altitude.     The    dotted 
liues    in  the   following   figures    represent    the    altitude. 


Fig. I 


Fir/.  2 


.,       y  Equilateral         Isosceles 


by  S> 

Named    \  Equiangular 
by  Angles  J    Acute-angled 


Fig.  3  Fig.  i 

Scalene 
Acute-angled    Right-angled  Obtuse-angled 


28.  A  Rig'ht-ang'led  Triangle  is  a  triangle  having 
one  right  angle. 

29.  The  Hypothenuse  of  a  right-angled  triangle  is 
the  side  opposite  the  right  angle. 

30.  An  Equilateral  Triangle  has  its  three  sides 
equal. 

31.  An  Isosceles  Triangle  has  two  of  its  sides 
equal. 

32.  A  Scalene  Triangle  has  no  two  sides  equal. 

33.  An  Acute -angled  Triangle  has  three  acute 
angles. 

34.  An  Obtuse-angled  Triangle  has  one  obtuse 
angle. 

35.  An  Equiangular  Triangle  has  its  three  angles 
equal. 

Rule  I.  To  find  the  area  of  a  triangle  when  the  base 
and  altitude  are  given:  Multiply  the  ba.se  by  one -half 
the    altitude. 


TUIANGLES  7 

Rule  II.  To  find  the  area  of  a  triangle  when  the 
three  sides  are  given  :  ISubtnict  each  side  sejmrately  from 
half  the  sum  of  ike  three  sides;  multij)ly  the  continued 
jyroduct  of  these  three  remainders  bij  the  half -sum  ; 
the    square    root    of    the  product  is  the  area. 

Rule  III.  To  find  one  dimension  of  a  triangle  when 
the  area  and  the  other  dimension  are  given:  Divide  the 
area  by  one -half  the  given  dimension. 

PROBLEMS 

1.  A  triangle  has  a  base  of  60  feet,  and  an  altitude 
of  20  feet ;   how  many  square  feet  does  it  contain  ? 

Ans.  600  sq.  ft. 

2.  Find  the  area  of  a  triangle,  the  length  of  whose 
base  is  50  feet,  and  whose  altitude  is  12  feet  4  inches. 

Ans.  308i  sq.  ft. 

3.  How  many  acres  in  a  triangular  lot  whose  base  is 
28  rods  and  altitude  18  rods?  Ans.  1  A.  92  sq.  rds. 

4.  What  is  the  cost  of  a  triangular  piece  of  land 
whose  base  is  30.96  chains  and  altitude  4.835  chains,  at 
$120  an  acre?  Ans.  $898.15—. 

5.  Find  the  area  of  a  triangle  whose  sides  are  respec- 
tively 60,  80,  and  100  feet.  Ans.  2,400  sq.  ft. 

6.  Find  the  base  of  a  triangle  whose  area  is  40  acres 
and  altitude  80  rods.  Ans.  160  rds. 

7.  A  triangle  contains  48  square  feet  63  square 
inches ;  the  base  is  12  feet  6  inches ;  required  the 
altitude.  Ans.  7  ft.  9  in. 

8.  Find  the  cost  of  sodding  a  triangular  plot  of 
ground  whose  sides  are  respectively  60,  75,  and  80  feet, 
at  15  cents  per  square  yard.  Ans.  $35.60+. 


8  M  EX  DURATION 

9.  Mr.  A  built  a  barn  in  the  form  of  the  letter  L  ; 
the  width  of  the  barn  at  each  end  is  60  feet,  and  the 
ridge  of  the  roof  is  24  feet  higher  than  the  foot  of  the 
rafters  ;  how  many  sqnare  feet  of  boards  were  required 
to  cover  the  three  gables?  Ans.  2,160  sq.  ft. 

10.  Owing  to  the  difference  in  the  soil  of  a  farmer's 
field,  which  is  square,  he  decided  to  plant  part  in  corn 
and  part  in  potatoes.  He  first  made  a  straight  furrow 
from  the  northeast  corner  of  the  field  to  the  southwest 
corner,  and  found  it  to  be  88  rods  long ;  from  the 
middle  of  this  furrow  to  the  southeast  corner  of  the 
field  is  44  rods.  He  planted  all  to  the  eastern  side  of 
the  first  furrow  drawn  in  potatoes  and  the  balance  in 
corn.  How  many  acres  in  the  field"?  How  many  acres 
did  he  plant  in  potatoes  ?    In  corn  ? 

Ans.  field,  24i  A.;   potatoes,  12to  A.;   corn,  12to  A. 

11.  Find  the  area  of  the  penta- 
gon as  shown  in  the  accompany- 
ing diagram.  The  diagonal  AC  is 
48  inches,  and  the  diagonal  AD,  36 
inches  ;  the  altitude  of  the  triangle 
ABC  is  16  inches,  of  the  triangle 
ACD,  20  inches,  and  of  the  triangle 
AED,  12   inches.     Ans.  1,080  sq.  in. 

THE  RIGHT-ANGLED  TRIANGLE 

36.  The  following  principles  relating  to  right-angled 
triangles  have  been  proved  by  Geometry.  An  examination 
of  Figures  5,  6,  and  7  will  be  helpful  to  the  student  in 
fixing  them  in  mind. 

Principle  I.  The  square  of  the  hijpothenuse  is  equal 
to  the  sum  of  the  squares  of  the  other  two  sides. 


EIOHT-AXGLKD    TRIAXGLES 


9 


\X^-'=:^ 

A    ^' 

io 

\| 

24 

10 

ciS 


Fig.  5 


Fig.  a 


Fig.  7 


Principle  II.  The  square  of  the  base,  or  of  the  per- 
pendicular, is  equal  to  the  square  of  the  hypothenuse 
diminished  by  the  square  of  the  other  side. 

Rule  I.  To  find  the  hypothenuse  :  Extract  the  square 
root  of  the  sum  of  the  squares  of  the  base  and  perpendicular. 

Rule  II.  To  find  the  base  or  perpendicular :  Extract 
the  square  root  of  the  difference  between  the  square  of  the 
hypothenuse  and  tli  square  of  the  given  side. 


PROBLEMS 

1.  What  is  the  length  of  the  hypothenuse  of  a  right- 
angled  triangle  (Fig.  5)  whose  base  is  4  feet,  and  whose 
perpendicular  is  3  feet ! 

Operation. —  "i/4"^  -f-  32  =  5  feet.    Ans. 

In  a  right-angled  triangle,  given  :  .... 

2.  The  base  5  feet,  perpendicular  12  feet,  to  find  the 
hypothenuse.  Ans.  13  ft. 

3.  The  base  8  inches,  perpendicular  15  inches,  to  find 
the  hypothenuse.  Ans.  17  in. 

4.  The  perpendicular  20  rods,  the  base  21  rods,  to  find 
the  hypotheuusc.  Ans.  29  rds. 


10  MENSURATION 

0.    The  perpendicular  3  yards,  the  base  4  yards,  to  find 
the  hypothenuse.  Ans.  5  yds. 

6.  The  hypothenuse  10  feet,   the  base  8  feet,  to  find 
the  perpendicular.  Ans.  6  ft. 

7.  The  hj-pothenuse  78  feet,  the  perpendicular  30  feet, 
to  find  the  base.  Ans.  72  ft. 

8.  Find  the  diagonal  of  a  square  whose  side  is  20  feet. 

Ans.  28.28+  ft. 

9.  Find  the  diagonal  of  a  cube  whose  edge  is  20  feet. 

Ans.  34.64+  ft. 

10.  The  gable  end  of  a  house  50  feet  wide  is  14  feet 
high;  what  is  the  length  of  the  rafters  ?     Ans.  28.65+  ft. 

11.  A  flag -pole  120  feet  high  casts  a  shadow  110  feet 
in  length ;  required  the  distance  from  the  end  of  the 
shadow  to  the  top  of  the  pole.  Ans.  162.78+  ft. 

12.  A  hunter  stood  40  feet  from  the  foot  of  a  tree 
and  shot  a  squirrel  on  the  top  of  a  tree  55  feet  high  ; 
how  far  did  the  bullet  go  before  hitting  the  squirrel, 
provided  the  starting  point  of  the  bullet  was  5  feet 
from  the  ground?  Ans.  64.03+  ft. 

13.  Thomas  Lighty,  a  carpenter,  directed  his  appren- 
tice to  cut  a  brace  of  the  proper  length  to  secure  a  per- 
pendicular in  a  sill,  bj^  measuring  6  feet  each  way  from 
the  corner  where  the  two  sticks  joined;  how  long  was 
the  braced  Ans.  8.48+  ft. 

14.  A  deiTick  at  a  granite  quarry  had  an  upright 
48  feet  long,  and  a  horizontal  ar^ji  2Q  feet  Jong ;  a  rope 
ran  on  pulleys  from  the  foot  of  the  perpendicular  to  the 
extremity  of  the  arm  and  thence  to  the  ground  ;  what 
length  of  rope  was  required,  no  allowance  being  made 
fi/r  fastening  the  rope?  Aus   100  ft. 


Q  UA  DRILA  TERALS  1 1 

15.  A's  horse  travels  a  mile  in  3  minutes,  and  B*s  a 
mile  in  4  minutes  ;  A  drives  at  this  rate  due  north  for 
24  minutes,  and  B,  starting  at  the  same  place,  due  east 
for  the  same  time  ;  how  far  are  they  apart  after  driving 
24  minutes  f  After  driving  at  the  same  rate  and  same 
direction,  respectively,  1  hour?  Ans.  10  mi.;   25  mi. 

16.  Williamson  Porter  in  hunting  bees  placed  his  bait 
on  a  stump  5  feet  high;  one  bee  flew  in  a  "bee  line" 
from  the  bait  75  feet  to  a  tree  due  west  from  the  stump, 
alighting  45  feet  from  the  foot  of  the  tree  ;  another  flew 
80  feet  east  to  a  tree,  alighting  65  feet  from  the  foot  of 
the  tree ;  if  the  ground  is  level,  and  both  trees  per- 
pendicular, what  is  the  shortest  distance  between  the 
trees?  Ans.  116.35-f  ft. 

17.  Mr.  A,  a  politician,  desired  his  three  neighbors, 
B,  C,  and  D,  to  vote  at  a  certain  election.  A  lived  100 
rods  south;  B,  120  rods  east;  C,  140  rods  north;  and  D, 
150  rods  west  of  the  polling  place.  A,  with  his  carriage, 
drove  from  home  in  a  straight  line  for  B,  thence  in  a 
straight  line  for  C,  and  thence  in  a  straight  line  for  D  ; 
from  D's  residence  he  drove  in  a  straight  line  to  the 
polling  place,  and  thence  in  a  straight  line  home  :  how 
far  did  he  drive,  provided  he  allowed  his  neighbors 
to  walk  home?  Ans.  2  mi.  155.77+  rds. 


QUADRILATERALS 

37.  A  QuadrHateral  is  a  portion  of  a  plane  bounded 
by  four  straight  Hues. 

38.  A  Parallelogram  is  any  plane  figure  of    four 
straight  sides,  the  opposite  ones  of  which  are  parallel. 

39.  A  Rectang'le  is  any  parallelogram  having  all  its 
angles  right  angles. 


12  MENSURATION 

40.  A   Square   is   a  rectangle   all   ot*  whose  sides  are 
equal. 

41.  A  Rhombus    is    a   parallelogram    having   all    its 
sides  equal  and  all  its  angles  oblique. 

42.  A   Rhomboid    is   a  parallelogram  whose  opposite 
sides  only  are  equal  and  whose  angles  are  oblique. 


Square        Eectangle       Rhoaibus         Rhomboid  Trapezoid  Trapezium 

r  1.  Parallelograms     ■{ 


r  1.  Square 
*    2.  Rectangle 


(3.  Rhombiis 
4.  Rhomboid 
(_  3    Ti-apezium 

43.  A  Trapezoid  is  a  quadrilateral  which  has  only 
two  of  its  sides  parallel. 

44.  A  Trapezium  is  a  quadrilateral  having  no  sides 
parallel. 

45.  The  Lower  Base  of  a  parallelogram  is  the  side 
upon  which  it  stands.  The  Upper  Base  is  the  side  oppo- 
site the  lower  base. 

46.  The  Altitude  of  a  parallelogram  or  of  a  trapezoid 
is  the  perpendicular  distance  between  two  bases.  The 
dotted  lines  in  the  figures  represent  the  altitude. 

47.  The  Diag'Onal  of  a  quadrilateral  is  a  straight  line 
joining  two  opposite  vertices. 

•  Rule    I.     To  find  the- area  of  any  parallelogram  :   Muh 
fiphj  the  base  hy  the  altitude.  .  . 

Rule  II.  To  find  the  area  of  a  trapezoid  :  MuUij)^ 
half  the  sum  of  the  pfirallel  sides  hf/  Ute  altitude. 


or  A  I>li  I  LA  TJi'JU  L  S  1 3 

Rule  III.  To  find  the  area  of  a  trapezium  :  Multiply 
the  diagonal  hij  half  the  sum  of  the  perpend iculai^s  drawn 
to  it  from  the  vertices  of  the  op2)Osite  angles. 

Note. — The  area  of  any  polygon  may  be  found  by  dividing  it 
into  triangles  and  finding  the  sum  of  their  areas. 

PROBLEMS 

1 .  The  side  of  a  square  is  20  feet ;    what  is  its  area  ? 

Ans.  400  sq.  ft. 

2.  A  has  a  rectanguhir  field  80  rods  long  and  GO  rods 
wide;   how  many  acres  does  it  contain?  Ans.  30  A. 

3.  Find  the  area  of  a  parallelogram  whose  base  is  150 
feet  and  altitude  80.5  feet.  Ans.  12,075  sq.  ft. 

4.  Find  the  area  of  a  rhombus  whose  base  is  20  feet 
and  altitude  6  feet  3  inches.  Ans.  125  sq.  ft. 

5.  A  field  in  the  form  of  a  rhomboid  has  a  base  136 
rods  long  and  an  altitude  of  550  yards  ;  hoAv  many  acres 
does  it  contain?  Ans.  85  A. 

6.  The  parallel  sides  of  a  trapezoid  are  70  rods  and 
90  rods  and  the  altitude  20  rods ;  find  the  area  in 
acres.  Ans.  10  A. 

7.  Find  the  area  of  a  trapezium,  a  diagonal  of  which 
is  50  feet,  and  the  perpendiculars  of  which  to  this 
diagonal  are   10   feet  and   35  feet.  Ans.  1,125  sq.  ft. 

8.  Required  the  area  of  a  trapezium,  the  lengths  of 
whose  sides  are  respectively  20,  30,  25,  nnd  35  chains, 
and  the  diagonal  40  chains,  making  two  triangles  whose 
sides  are  respectively  35,  20,  and  40;   and  30,  25,  and  40. 

Ans.  72  A.  71  sq.  rds.  17  sq.  yds.  5  sq.  ft.  143.136  sq.  in, 

9.  Required  the  area  of  a  regular  hexagon  whose  side 
is  6  inches.  Ans.  93.528+  sq.  in. 

Suggestion.  — Divide  into  6  equilateral  triangles,  or  3  rhom- 
buses, or  2  trapezoids. 


14 


MENS  UH  ATI  ON 


THE  CIRCLE 

48.  A  Circle  is  a  plane  figure  bounded  by  a  curved 
line,  every  point  of  which  is  equally  distant  from  a  point 
within  called  the  center. 


C\tc«ni£e^^^ 


Circle 
{Fig.  1) 


The  Circle 


Clrcumscribecl  Square 


^    ~^ 

Inscribed 

\ 
J 

Scfiiare 

^    ^ 

{Fig.  2) 


Lines 


Curved 


Straight  . 


Circumference 
Arc 

1.  Diameter 

2.  Radius 

3.  Secant 

4.  Tangent 

5.  Claord 

f  1.  Semi-circle 

Surfaces  of  portions  of  Circle  J  "'  ^^^  ^  ^ 
I  3.  Sector 

4.  Segment 


49.  The  Circumference  of  a  circle  is  the  line  which 
bounds  the  circle. 

50.  An  Arc  is  any  portion  of  the  circumference. 

51.  The  Diameter  of  a  circle  is  a  straight  line  passing 
through  the  center,  and  having  its  extremities  in  the  cir- 
cumference. 

52.  The  Radius  of  a  circle  is  a  straight  line  drawn 
from  the  center  to  any  part  of  the  circumference.  It  is 
one -half  the  diameter. 


HIE    CIliCLK  15 

63.  A  Secant  is  a  straight  line  which  cuts  the  cir- 
cumfereuce  at  two  points. 

54.  A  Tangfent  is  a  straight  line  which  touches  the 
circumference,  but  does  not  cut  it. 

55.  A  Chord  is  a  straight  line  joining  the  extremities 
of  an  arc. 

56.  A  Semi-circle  is  half  a  circle. 

57.  A  Quadrant  is  a  quarter  of  a  circle. 

58.  A  Sector  of  a  circle  is  a  portion  of  a  circle 
included  by  two  radii  and  the  arc  intercepted  by  them. 

59.  A  Seg'ment  of  a  circle  is  a  portion  of  a  circle 
included  between  an  arc  and  its  chord. 

60.  The  ratio  of  the  circumference  to  the  diameter  is 
the  same  for  all  circles.  For  convenience  tiiis  ratio  is 
represented  by  the  Greek  letter  tt  (pi).  The  approximate 
numerical  value  of  ^  is  3.1416,  which  is  the  value  used 
in  this  work. 

Rule  I.  To  find  the  circumference  of  a  circle  when  the 
diameter  is   given  :     Multiply  the  diameter  by  3.1416. 

Rule  II.  To  find  the  diameter  of  a  circle  when  the  cir- 
cumference is  given:     Divide  the  circumference  hy  3.1416. 

Rule  III.  To  find  the  area  of  a  circle  when  its  diameter, 
radius,  and  circumference  are  given,  or  when  either  is 
given  : 

(a)  Multiply  the  circumference  hy  one- fourth  of  its 
diameter ;    or, 

(b)  Multipthj  the  square  of  the  diameter  hy  .7854;  or, 

(c)  Multiply  the  square  of  the  radius  hy  3.1416;    or, 

(d)  Multiply  half  the  distance  roimd  hy  half  the  dis 
tance  through. 


Rule  IV.     To  find  either  dimension,  when  the  area  is 
given  : 

Let  A  represent  the  area  of  a  circle,  C  the  circumference,  D  the 
diameter,  and  K  the  radius  ;  then. 


I>-Vw-:-.-.',    0=(T/:y^)X  3.1416;    R  =  i/^-, 
Rules  I,  II,  and  III  may  be  stated  as  follows  : 

C  =  7rD   or  27rR  ;    D=-^  ;    A  =  tt  R2  or   ^. 

Rule  V.  To  find  the  side  of  an  inscribed  square  when 
the  diameter  is  given  : 

(a)  Extract  the  square  root  of  one -half  the  square  of 
the  diameter;  or, 

(b)  Multiply  the  diameter  hy  .7071. 

Rule  VI.  To  find  the  area  of  a  circular  ring  formed  bj'- 
two  concentric  cii'cles:  Find  the  areas  of  the  circles 
separately  and  tal'e  their   difference. 


PROBLEMS 

1.  Find  the  circumference  of  a  circle  whose  diameter 
is  20  inches.  Ans.  62.832  in. 

2.  Find  the  circumference  of  a  circle  whose  diameter 
is  39  feet.  Ans.  122.522  ft. 

3.  Find  the  diameter  of  a  circle  whose  circumference 
is  128.8056  yards.  Ans.  41  yds. 

4.  Find  the   diameter  of  a  circle  whose  circumference 
is  37.6992  inches.  Ans.  12  in. 

5.  Find  the  radius  of  a  circle  whose  circumference  is 
119.3808  inches.  Ans.  19  in. 

6.  Find   the   radius    of   a   circle  whose   circumference 
is   100.5312  rods.  Ans.  16  rds. 

7.  How  far  is  it  round  a  tree  which  measures  4    feet 
over  the  stump?  Ans.  12.5664  ft. 


THE    ClKCLi:  17 

8.  The  spokes  of  tlie  fore  wheel  of  a  buggy  are 
2  feet  long,  and  of  the  hind  wheel  2  feet  2  inches  ;  no 
allowance  being  made  for  the  thickness  of  the  hub,  how 
much  longer  is  the  tire  of  the  hind  wheel  than  of  the 
fore  wheel?  Ans.  1.0472  ft. 

9.  Four    flower -pots   each   with    a   lower   diameter  of 

1  foot  were  placed  on  a  bench  10  feet  long  and  16 
inches  wide ;  how  much  of  the  bench  remained  un- 
covered? Ans.  1467.6096  sq.  in. 

10.  John  threw  a  base -ball  2  inches  in  diameter  at  a 
paste -board  card  2  feet  square,  and  made  in  the  card 
15  holes  of  the  same  circumference  as  the  ball ;  how 
much  of  the  surface  of  the  card  remained  unbroken  ? 

Ans.  528.876  sq.  in. 

11.  Certain  trustees  asked  for  bids  to  fresco  the  ceil- 
ing of  their  church,  which  is  60  feet  long  and  40  feet 
wide.  A  contractor  offered  to  place  on  the  ceiling  8 
circular  designs  12  feet  in  circumference  at  $3  each,  and 
paint  the  remaining  surface  at  9  cents  per  square  j-ard, 
or  to  place  on  the  ceiling  8  square  designs,  the  perimeter 
of  each  to  be  12  feet,  at  $3  each,  and  paint  the  remain- 
ing surface  at  1  cent  per  square  foot.  The  trustees 
chose  the  second  offer.  How  much  did  it  cost  them ! 
How  much  did  they  save  or  lose  by  taking  the  second 
offer  instead  of  the  first?        Ans.  cost,  $47.28;  lost,  20c. 

12.  The  minute  hand  of  a  town  clock  is  2  feet  long ; 
over  how  much  surface  does  it  pass  in  20  minutes  ? 
In   15  minutes?  Ans.  4.1888  sq.  ft.;   3.1416  sq.  ft. 

13.  How    many   times    will    a   wheel    whose   radius    is 

2  feet  revolve  in  going  1  mile?        Ans.  420.168+  times. 

14.  Find  the  width  of  the  ring  between  two  concen- 
tric circles  whose  circumferences  are  respectively  15.708 
and  31.416  feet.  Ans.  2^  ft. 

B 


IS  MENSV  RATION 

15.  Find   the  area  of   a  circle  inscribed   in   a   square 
containing  576  square  feet.  Ans.  452.39+  sq.  it. 

16.  Find   the  area  of   a  circle  circumscribed   about  a 
square  containing  256  square  feet.      Ans.  402.109+  sq.ft. 

17.  The    diameter  of   a   cii'cle    is    100    feet ;    find   the 
side  of   the  inscribed  square.  Ans.  70.71  ft. 

18.  The  circumference  of  a  circle  is  62.832  feet;   find 
the  side  of  an  inscribed  square.  Ans,  14.142  ft. 

19.  The   area  of   a    circle  is   1,963.44  square   inches ; 
find  the  side  of  an  inscribed  square.  Ans.  35.355  in. 

THE   ELLIPSE 

61,  An  Ellipse  is  a  curved  plane  formed  by  an 
oblique  section  of  either  a  cone  or  a  cylinder,  passing 
through  its  curved  surface,  but  not 
touching  its  base;  the  sum  of  the  dis- 
tances from  every  point  of  the  bound- 
ing line  to  two  fixed  points,  called 
foci,  is  equal  to  a  line  drawn  through 
these    points    and    terminated    by   the  Ellipse 

curve.     The    line    drawn    through   the 
foci   is    a   transverse    axis ;    a   perpendicular  to  this   axis 
at  its  middle  point  is  the  conjugate  axis. 

Rule.     To  find  the  area  of  an  ellipse  :   Find  the  product 
of  the  semi -axes,  a7id  multiply  this  iwoduct  by  3.1416. 

PROBLEMS 

1.   What   is    the   area  of   an    ellipse  whose    transverse 
axis  is  40  inches  and  conjugate  axis  32  inches! 

Ans.  1,005.312  sq.  in.. 


SIMILAH    rLAXK    FIG  V  RES  19 

2.  The  axes  of  an  elliptical  flower  bed  are  20  and 
16  feet;    what  is  its  area?  Ans.  251.328  sq.  ft. 

3.  The  transverse  axis  of  an  elliptical  fish  pond  is 
80  feet  and  conjugate  axis  60  feet ;    find  the  area. 

Ans.  3,769.92  sq.  ft. 

4.  How  much  larger  is  a  circle  whose  diameter  is 
10  inches  than  an  ellipse  whose  transverse  axis  is  equal 
to  the  diameter  of  the  circle,  and  conjugate  axis  f  of 
the  length  of  the  transverse  axis?        Ans.  31.416  sq.  in. 


SIMILAR   PLANE  FIGURES 

62.  Similar  Plane  Fig-upes  are  those  which  have  the 
same  form.  Their  angles  are  equal  and  their  like  dimen- 
sions proportional.  Squares,  equiangular  triangles,  regu- 
lar polygons  of  the  same  number  of  sides,  and  circles 
are  similar.  The  surfaces  of  similar  solids  are  also 
proportional.  The  like  dimensions  of  circles  are  their 
radii,  diameters,  and  circumferences,  and  of  other  plane 
figures,  those  which  are  like  placed.  The  following 
principles    are    derived   from    Geometry : 

Principle  I.  The  areas  of  similar  surfaces  are  to 
each   other  as    the   squares   of  their  like   dimensions. 

Principle  II.  The  like  dimensions  of  similar  sur- 
faces are  to  each  other  as  the  square  roots  of  their  areas. 

PROBLEMS 

1.  How  many  circles  eacli  8  inches  in  diameter  will 
equal  the  area  of  a  circle  4  feet  in  diameter?       Ans.  36. 

2.  John's  slate  is  14  by  10  inches,  and  William's  7  by 
5  inches  ;   how  do  they  compare  in  size  ? 

Ans.  John's  =  4  times  William's. 


20  MENSURATloyf 

3.  Two  farms  of  exactly  the  same  shape  contain 
respectively  144  and  169  acres.  One  side  of  the  former 
is  480  rods ;  required  the  corresponding  side  of  the 
latter.  Ans.  520  rds. 

4.  A  has  a  square  field,  the  side  of  which  is  40  rods, 
worth  $50  an  acre,  and  B  a  square  field,  i  as  long,  worth 
$100  an  acre ;  which  is  the  more  valuable,  and  how 
much?  Ans.  A's,  $250  more. 

5.  If  it  cost  $40  to  (-arpet  a  rectangular  room  20  feet 
long,  how  much  will  it  cost  to  carpet  a  room  of  similar 
shape  25  feet  long?  Ans.  $62.50. 

6.  A  lady  has  two  circular  flower  beds  ;  the  one  62  feet 
in  diameter,  the  other  Si  feet ;  the  first  is  how  many  times 
the  size  of  the  second?  Ans.  4  times. 

7.  The  area  of  a  rectangular  field  is  2,880  square  rods  ; 
its  sides  are  as  4  to  5  ;   what  is  the  length  of  each  side  ? 

Ans.  48  rds.;   60  rds. 

8.  The  hypothenuse  of  a  right-angled  triangle  is  25 
inches  ;  what  is  the  hypothenuse  of  a  similar  triangle 
which  contains  twice  the  area?  Ans.  35.35+  in. 

9.  The  area  of  a  trapezoid  is  216  square  rods,  and  its 
altitude  12  rods  ;  find  the  altitude  of  a  similar  trapezoid 
whose  area  is  384  square  rods.  Ans.  16  rds. 

10.  A  battle  ship  whose  anchor  weighs  4,000  pounds 
requires  a  cable  3  inches  in  diameter  ;  what  should  be  the 
diameter  of  the  cable  when  the  anchor  weighs  4  tons? 

Ans.  4.242+  in. 

11.  If  a  cistern  can  be  filled  hy  a  pipe  3  inches  in 
diameter  in  24  minutes,  in  what  time  can  it  be  filled  by 
a  pipe  9  inches  in  diameter  ?  Ans.  2  min.  40  sec. 

12.  How  many  saucers  each  5  inches  in  diameter  will 
equal  in  area  the  surface  of  a  round  table  5  feet  in 
diameter  ?  Ans.  144. 


SIMILAR    PLANU    FIGURES  21 

13.  If  the  circumference  of  a  circle  is  2  feet,  what  is 
the  circumference  of  a  circle  Sro  times  as  great  ? 

Ans.  3^  ft. 

14.  The  altitudes  of  two  similar  triangles  are  21  feet 
and  5i  feet ;   what  is  the  relation  of  their  areas  ?    Ans.  16. 

15.  The  area  of  a  triangle  is  5,400  square  feet,  and  its 
sides  are  proportional  to  the '  numbers  9,  12,  and  15; 
required  the  length  of  the  sides. 

Ans.  90  ft.,  120  ft.,  150  ft. 

16.  Mr.  Bowers  wished  to  sow  clover  seed  on  two 
fields,  one  40  rods  square  and  the  other  60  rods  square; 
he  estimated  that  one  bushel  would  be  sufficient  for  the 
smaller  field  ;  what  should  have  been  his  estimate  for 
the  larger  field?  Ans.  2i  bu. 

17.  Sarah  baked  for  John  2  buckwheat  cakes  each  6 
inches  in  diameter,  and  for  Rufus  8  cakes  of  the  same 
thickness  each  3  inches  in  diameter;  John  accused  Sarah 
of  unfairness  ;    was  she  unfair  ? 

18.  If  it  requires  96  square  inches  of  tissue  paper  to 
cover  a  cube  whose  edge  is  4  inches,  how  much  would  be 
required  to  cover  a  cube  whose  edge  is  8  inches  ? 

Ans.  384  sq.  in. 

19.  It  cost  Mr.  McDowell  $80  to  pave  his  walk,  which 
is  6  feet  wide  and  100  feet  long  ;  Mr.  Blair  paid  $320 
for  paving  his  walk  similar  in  shape  to  Mr.  McDowell's, 
with  the  same  material;  what  are  the  dimensions  of  Mr. 
Blairls  walk?  Ans.   12  ft.  wide;   200  ft.  long. 


MENSURATION    OF    SOLIDS 

PRISMS   AND    CYLINDERS 

63.  A  Solid  or  Body  has  three  dimensions, —  length, 
breadth,  and  thickness.  The  bounding  planes  are  its 
faces,  and  their  intersecting  lines  its  edges. 

64.  A  Prism  is  a  solid  whose  two  ends  are  parallel, 
similar,  and  equal,  and  whose  sides  are  parallelograms. 

They  are  named  according  to  the  form  of  their  bases; 
as  triangular,  quadrangular,  pentagonal,  hexagonal,  etc. 

65.  The  Altitude  of  a  prism  is  the  perpendicular  dis- 
tance between  its  bases.  In  Figs.  4,  5,  and  6  the  per- 
pendicular dotted  lines  represent  the   altitude. 

66.  A  Rig'ht  Prism  is  a  prism  whose  lateral  edges 
are  perpendicular  to  the  bases. 

67.  An  Oblique  Prism  is  a  prism  whose  lateral  edges 
are  not  perpendicular  to  the  bases. 


Right  Prisms 


Oblique  Prisms 


Frustums  of  Prisms 


V 


w 


a 


t^ 


Q 


Fig.  1    Fig.  2  Fig.  3     Fig.  i        Fig.  5        Fig.  6         J'lg. 


■  U 

Fig.  8  Fig. 


68.  A  Frustum  of  a  prism  is  that  part  which  remains 
after  cutting  olf  a  part  of  it  by  passing  a  plane  through 
it  not  parallel  to  either  base.  This  is  also  called  a 
trunmUd  prism. 

(22) 


MENSURATION    OF   SOLIDS  23 

69.  A   Parallelepiped   is   a  prism    bounded    by   six 
parallelograms,  the  opposite  faces  being  parallel. 

70.  A  Rig'ht    Parallelepiped    is   one   whose    lateral 
edges  are  perpendicular  to  its  bases. 


Rectangular 

Parallelopipeds 
Cuhc                                   Rhomb 

Rhombic  Prism 

\ 

\ 

^ 

N.         -^      .^i 

a 

1 

■/  /^\ 

a 

\ 

k 

Right 
Fig.  10 

Right 
Fig. 11 

Ol.lique 
Fig.  12 

Oblique 
Fig.  Vi 

71.  A  Rectang-ular  Parallelepiped  is  a  right  paral- 
lelopiped  whose  bases  are  rectangles.  All  of  its  faces 
are    rectangles. 

72.  A  Cube  is  a  parallelopined  whose  faces  are  all 
equal    squares. 

73.  A  Rhomb  is  a  parallelopiped  whose  faces  are  all 
equal  rhombuses. 

74.  A  Rhombic  Prism  is  a  parallelopiped,  whose 
faces  are  rhombuses  or  rhomboids,  having  each  pair  of 
opposite  faces  equal,  but  not  all  its  faces  equal.  In  each 
figure  above,  a  represents  the  altitude. 

75.  A  Cylinder  is  a  solid  bounded  hy  a  uniformly 
curved  surface,  and  having  for  its  ends  two  equal  par- 
allel circles.  Any  section  of  a  cylinder  parallel  to  the 
ends  is  a  circle  equal  to  either  end. 

76.  A  Right  Cylinder  is  one  whose  ends  are  per- 
pendicular to  its  sides. 

77.  An  Oblique  Cylinder  is  one  whose  ends  are  not 
perpendicular  to  its  ^ides. 


24 


MENSURATION 


78.  The  Bases  of  a  cylinder  are   the   two  equal   and 
parallel  circles. 

79.  The  Altitude  of  any  cylinder  is  the  perpendicu- 
lar distance  between'  the  planes  of  its  bases. 

80.  The    Axis    of    a    cylinder   is    a    line    joining    the 
centers  of   its  bases. 

81.  A  Cylindroid   is  a  solid  resembling  a  cylinder, 
but  having  its  bases  elliptical. 


Fia.  L4 


Oblique  Cyliader 


Mff.  15 


Cylindroid 


Fig 


The  right  cylinder  (Fig.  14)  may  be  conceived  to  be 
generated  by  the  revolution  of  the  rectangle  ABCD 
about   its   side   BD,  as   an   axis. 

A  rigJit  section  of  a  prism  or  a  cylinder  is  a  sec- 
tion made  by  a  plane  perpendicular  to  the  axis ;  as  X 
(Fig.  15). 


0iM 


FigAI 


m 


Fig.\^ 


Fig.l^ 


82.  By  examining  Figs.  17,  18,  and  19,  it  will  be 
observed  that  the  convex  surface  of  each  maj'  be  represented 
by  a  rectangle,  the  perimeter  of  the  prism  or  cylinder  being 


SURFACES    OF    miSMS   AND    CYLINDERS  25 

equal  to  the  base  of  the  rectangle,  and  the  altitude  of  the 
prism  or  cylinder  equal  to  the  altitude  of  the  rectangle  ; 
hence  the  following  rule  : 

Rule.  To  find  the  convex  surface  of  a  right  prism  or 
right  cylinder  :  Multiply  the  j^erimeter  of  the  base  by  the 
altitude. 

For  the  entire  surface,  add  to  the  convex  surface  the 
areas  of  the  two  bases. 

The  convex  surface  of  any  prism  may  be  found  by 
adding  together  the  areas  of  the  lateral  faces. 

The  convex  surface  of  any  cylinder  may  be  found  by 
finding  the  product  of  the  perimeter  of  a  right  section 
of   the  cvlinder  and  the   axis. 


SURFACES  OF  PRISMS  AND  CYLINDERS 

PROBLEMS 

1.  What  is  the  convex  surface  of  a  prism  whose  alti- 
tude is  10  feet  and  perimeter  of  the  base  20  feet  ? 

Ans.  200  sq.  ft. 

2.  What  is  the  convex  surface  of  a  prism  whose  alti- 
tude is  10  yards  and  perimeter  of  the  base  10  yards  ? 

Ans.  100  sq.  yds. 

3.  What  is  the  convex  surface  of  a  hexagonal  prism 
whose  altitude  is  20  feet  and  each  side  of  the  base  12 
inches?  Ans.  120  sq.  ft. 

4.  What  is  the  convex  surface  of  a  cylinder,  the  circum- 
ference of  whose  base  is  32.66  feet  and  altitude  10  feet  ? 

Ans.  326.6  sq.  ft. 

5.  What  is  the  convex  surface  of  a  cylinder  whose  alti- 
tude is  20  feet  and  the  radius  of  the  base  5  feet  ? 

Ans.  628.32  sq.  ft. 


26  MEXSURATION 

6.  The  diameter  of  the  base  of  a  ej'linder  is  20  feet 
and  the  altitude  30  feet  ;   required  the  convex  surface. 

Aus.  1,884.96  sq.  ft. 

7.  Find  the  entire  surface  of  a  quadrangular  prism 
whose  altitude  is  20  feet  and  each  side  of  the  base  3  feet. 

Ans.  258  sq.  ft. 

8.  What  is  the  convex  surface  of  a  pentangular  prism 
whose  altitude  is  30  feet  and  each  side  of  the  base  4  feet? 

Ans.  600  sq.  ft. 

9.  What  is  the  convex  surface  of  a  cylinder  whose 
altitude  is  4  yards  and  diameter  of  the  base  3  feet  ? 

Ans.  113.0976  sq.  ft. 

10.  The  altitude  of  a  triangular  prism  is  5  inches  and 
each  side  of  the  base  6  inches  ;  what  is  the  entire  sur- 
face f  Ans.  121.1768  sq.  in. 

11.  Required  the  entire  surface  of  a  parallelopiped  8 
feet  long,  5  feet  wide,  and  3  feet  high.       Ans.  158  sq.  ft. 

12.  What  is  the  entire  surface  of  a  cube  whose  edge  is 
7  feet "?  Ans.  294  sq.  ft. 

13.  What  is  the  entire  surface  of  a  rhomb,  each  face  of 
which  is  a  parallelogram,  whose  base  is  2  feet  2  inches  and 
altitude  2  feet?  Ans.  26  sq.  ft. 

14.  A  door  8  feet  high  and  4  feet  wide  revolves  around 
one  of  its  sides  as  an  axis  ;  what  is  the  convex  surface  of 
the  cylinder  generated  by  it  f  Ans.  201.0624  sq.  ft. 

15.  A  door  8  feet  long  and  4  feet  wide  revolves  upon  a 
point  in  its  center  ;  what  is  the  convex  surface  of  the 
cylinder  generated  by  it  ;   what  is  the  entire  surface  ? 

Ans.  100  5312  sq.  ft.;    125.664  sq.  ft. 

16.  The  diameter  of  the  base  of  a  cjdinder  is  16 
inches,  and  the  altitude  2o-  feet ;  how  many  square  feet 
ai-e  there  in  the  lateral  surface?  Aus.  10.472  sq.  ft. 


ILLUSTRATIVE    PROBLEMS  27 

ILLUSTRATIVE    PROBLEMS 

Solved  by  the  Prismoidal  Formula 

83.  The  special  feature  of  this  work  is  the  applica- 
tion of  the  Prismoidal  Formula  to  finding  the  volume  of 
solids.  Since  this  one  rule  covers  so  broad  a  field  of 
measurement,  it  is  thought  that  a  thorough  knowledge 
of  the  method  of  using  it  in  finding  the  volume  of  the 
cube,  cylinder,  etc.,  as  well  as  of  those  solids  for  which  the 
rules  usually  given  are  more  complex,  will  be  beneficial. 
The  author  does  not  advise  the  teacher  to  discard  the 
well  known  rules  for  finding  the  volume  of  a  prism, 
cylinder,  etc.;  but  he  does  recommend  that  this  rule  be 
substituted  for  those  usually  quoted  for  finding  the 
volume  of  many  other  solids,  as  the  frustum  of  a  pyra- 
mid or  of  a  cone.  Both  rules  are  given  under  each 
subject;  this  will  enable  the  teacher  in  each  case  to  take 
his  choice.  In  order  that  the  pupil  may  be  made  quite 
familiar  with  the  rule,  a  number  of  illustrative  problems 
are  solved,  and  such  suggestions  made  as  will  enable  the 
pupil  to  use  it  intelligently  in  following  chapters.  These 
are  arranged  in  a  separate  section  at  this  place  so  that 
comparisons  may  be  more  advantageously  made.  Future 
reference  is  made  to  this  chapter.  For  a  derivation  of 
the  formula,  etc.,  see   Supplement. 

Rule. — To  find  the  volume  of  a  prismoid,  etc.: 

84.  Multiply  the  sum  of  both  end  areas  and  four  times 
the  area  of  a  section  half  way  between  them^  by  one-si^h 
the  altitude. 

Letting  a  =  the  altitude  or  length  ;  B,  the  area  of  either  base, 
and,  when   unequal,  the   larger   base;    b,   the  area   of   the  opposite 


28 


MEXSURATWX 


base,  aud,  when  imequa],  the  smaller  ba.se;  and  4m,  fonr  times  the 
area  of  a  section  half  way  between  the  bases,  the  rule  may  be 
stated    as    follows:     |(B  +  b  +  4m). 

This  formula  will  be  used  throughout  this  work. 

In  all  figures  in  the  following  chapter,  the  middle  sec- 
tion is  represented  by  dotted  lines  and  the  altitude  is 
marked  with  the  lettei*  a. 


1.    Required   the  volume   of   a    cube 
whose  edge  is  6  feet. 

Operation.—  B  =  6  X  6  =  36  sq.  ft. 
b=6  X  6=36       " 
4m  =4  (6  X  6)  =  144  sq.  ft. 
a  =  6 
Substituting  in  formula,  -|(B  -|-  b  +  4m; 
we  have,  f  (36  +  36  +  144)=  216  cubic  feet,  volume 


/ 

b 

z 

6 

ni 

4 

/ 

B 

A 

Remark,— It  will  be  observed  that  in  all  uniform  solids  not 
tapering  or  having  an  apex,  such  as  prisms  and  cylinders,  the  area 
of  both  bases  and  of  once  the  middle  section  is  the  same  ;  this  neces- 
sitates only  one  calculation  for  these  areas. 


2.    Required  the  number  of  cubic  feet  of  air  in  a  room 
20  feet  long,  8  feet  wide,  and 
12  feet  high. 


Operation. — 

B=  20X8=  160  sq.  ft. 


b  =  20X8  =  160 

4m  =  4  (20X8)  =640  sq.  ft. 
a=12 


/ 

b 

'a 

y 

20 

a 

m 

'8 

-  20 

\i 

/ 

B 

A 

20 


Substituting  in  formula,  |  (B  +  b  +  4m) 

we  have,  V"  i  160  -f  160  -\~  640)=  1,920  cubic  feet. 


IL L I  'SIR A  TJ  I  K     I'JxUBLKMS 


29 


8.    What  is  the  vohniie  of   a  triaiigiihu'  prism  whose 
altitude  is  10  feet,  and  each  side  of  the  base 
4  feetf 

Operation.  —  B  =    6.928  sq.  ft.  (Art.  35,  R.  II) 
b=    G.928       "  "  *' 

4m  =  27.712       " 
a  =  10 
Substituting  in  formula,  |  (B  +  b  +  4m) 
we  have,  \^  (6. 928 +  6. 928  4-27. 712)  =  69.28  cubic 
feet,  volume. 


\,.  VI 

K  /41 

a 

■-.  m 

4/ 

10 

4^N 

/'4 

\    B 

4/ 

N 

/\ 

4.   What  is  the  cost  of   a   piece  of   timber   18  inches 
square   and  20  feet  long,  at  $.30  a  cubic  foot? 


18^ 


iPl 


18"'; 


18  B 


:i  20 

Operation.  —  18  inches  =  f  ft. 

B=tXf  =  f  sq.  ft. 
b=tXf=f       " 
4m  =  4  (IX  f)  =9  sq-  ft. 
a  =  20 
Substituting  in  formula,  |(B  +  1^  -f  4m) 
we  have,  -^  (1  +  t  4-  9)  =  45  cubic  feet,  volume. 


5.   What  is  the  volume  of  a  cylinder  whose  radius  is 
1  foot  3  inches,  and  altitude  30  feet? 

Operation. —  1  foot  3  inches  =  1  ft. 

B=    (I)  2  X  3.1416=    4.90875  sq.  ft.  (Art.  60,  R.  Ill  c) 

b=    (1)2X3.1416=   4.90875 
4m  =  4(f)  2X3.1416  =  19.635  "  " 

a  =  30 
Substituting  in  formula,  |  (B  +  b  +  4m) 
we  have,  ^8^(4.90875  +  4.90875  +  19.635)=147.2625  cubic  feet,  volume. 


30 


MEX.<L'liAnu\ 


G.    What  is  the  volume  of  a  pyramid  whose  base  is  6 
feet  square,  and  altitude  24  feet  ? 


Operation. —  In  the   triangle     X  Y  Z, 


1  =  0 


(5  _|_  0  -J-  2  or  3  is  the  length  of  the  line  RS, 
or  the  side  of  the  middle  section. 
B  =  6X6  =  36  sq.  ft. 

4m  =  4  (3X3)  =36  sq.  ft. 
a  =  24 
Substituting  in  formula,  |  (B  +  b  +  4m) 
we  have,   V- (36  +  0  +  36)  =  288  cubic 

feet,  volume. 

Remarks.— (a)  It  is  apparent  that  in  the 
pyramid  and  cone  the  area  of  the  smaller 

base  is  Ze:'o.  (b)  Since  similar  surfaces  are  to  each  other  as  the 
squares  of  their  like  dimensions,  and  since  in  the  pyramid  and  cone 
the  side  or  diameter  of  the  lower  base  is  twice  the  length  of  the  side 
or  diameter  of  the  middle  section,  it  follows  that  the  area  of  the  larger 
base  is  equal  to  four  times  the  area  of  the  middle  section.  There- 
fore, only  one  calculation  is  necessary  for  both  areas. 


7.  Find  the  volume  of  the  frustum  of  a  square  pyramid 
whose  altitude  is  40  feet,  each  side  of  the  lower  base  12 
feet,  and  of  the  upper  base  10  feet. 

Operation.— The  face  W  X  Y  Z,  is  a 
trapezoid.  Therefore,  i  the  sum  of  the 
parallel  sides,  or  i  (12  +  10)  or  11  feet  is 
the  length  of  the  line  RS,  the  side  of  the 
middle  section. 

B  =  12X12  =  144  sq.  ft. 

b  =  ioxio  =  ioo     *' 

4m  =  4  (11  X  11)  =  484  sq.  ft. 

a  =  40 
Substituting  in  formula,  f  (B  +  b  +  4m) 
we  have,  \^  (144  +  100  +  484)=  4,853^ 

jubic  feet,  volume. 
N.  B. —  By  this  rule  no  root  is  extracted. 


ILL  i'STRA  Tl  ]  E    moliLKMS 


31 


8.    Find  the  volume  of  a  eoiie,  the  diameter  of  whose 
base  is  20  feet,  and  altitude  36  feet. 


Operation. —  20  -(-0-^2  =  10,  diameter  of  middle 
section. 
B  ==  lO*''  X  3.141G    =  314. IG  sq.  ft,  (Art.  60,  R.  Ill  e) 
o-=0 
4m  =  4(52X3.U16)=314.16     " 

u  =  3G 
Substituting  in  formula,  f  (B  +  b  +  4m) 
we    have,  ^^  (314.16  +  0  +  314.16)  =  3,769.92   cubic 
feet,  volume. 


b=o 


\ 

lUA 

/^.,_ 

']>\ 

/        ^ 

_\ 

20         ^ 

9.  The  altitude  of  the  frustum  of 
a  cone  is  60  feet,  the  diameter  of  the 
lower  base  12  feet,  and  of  the  upper 
base  8  feet;    required  the  volume. 

Operation.-  12"+^  h-  2  =  10,    diameter    of 
middle    section. 
B  =  62X3. 1416  =  113. 0976  sq.ft. (Art. 60, R.IIIc) 
b  =  4^X3.1416=    50.26.56       "  *' 

4m  =  4(5-X3.1416)  =  314.1G       ''  ''         " 

a  =  60 
Substituting  in  formula,  |(B  -f-  b  -f-  4m) 
we    have,  ^i-  (113.0976  +  50.2656  +  314.16)'  = 
4,775.232  cubic  feet,  volume. 


From  the  principles  of  Geometrj'  we  have  the  follow- 
ing :  The  volume  of  the  frustum  of  a  cone  is  equal  to 
the  sum  of  the  volumes  of  three  cones,  having  for  a 
common  altitude  the  altitude  of  the  frustum,  and  for 
bases  the  two  bases  of  the  frustum  and  a  mean  propor- 
tional between  them.  This  involves  the  extraction  of 
a  root.     In  the  above   solution   no  root  is  extracted. 


32 


Mi!:ysuiiATJoy 


10.    Find  the  volume  of  a  sphere  whose  diameter  is  40 
inches. 


Operation.  —  In  the  sphere  any 
diameter  may  be  re-garded  as  the  alti- 
tude, and  the  area  of  any  great  circle,  the 
area  of  a  middle  section.  The  area  of 
each  base  is  Zero. 

b  =  0 
4m  =  4  (20^  X  3.1416)  =  5,026.56 
sq.ft.  (Art.  60,  R.  Ill  c) 
a  =  40 


trO 


BiO 


Substituting  in  formula,  |  (B  -(-  b  -f-  4m) 

we  have,  ^(0  +  0  +  5,026.56)=  33,510.4  cubic  inches,  volume. 


11.    What  is  the  volume  of  a  prolate  spheroid  whose 
shorter  axis   is  10  feet 
and     longer     axis     15 
feet '? 


Operation.—  Regard 
the  longer  axis  as  the  alti- 
tude, and   the  shorter  axis 
as  the  diameter  of  the  mid- 
dle   section.     The    area    of 
each  base  is  Zero. 
B  =  0 
b  =  0 
4in  =  4  (5^X3.1416)=  314.16  sq.  ft.  (Art.  60,  R.  Ill  c) 
a  =  15 
••Substituting  in  formula,  -|(B  +  b  +  4m) 
we  have,  -^  (0  +  0  +  314.16)  =  785.4  cubic  feet,  volume. 

Remark.— The  problem  can  also  be  solved  by  using  the  shorter 
axis  as  the  altitude,  and  finding  the  area  of  the  elliptical  middle 
section. 


ILL  isTHA  TI  f  L    LROBLKMS 


33 


12.    Required  the  volume  of  a  wedge,  the  dat  end  of 
which  is  a  rectaiio:le  4    by  8  inches, 
and  the  height  24  inches. 

Operation,— B  =  4  X  S  =  32  sq.  in. 
b  =  0 
4m  =  4  (2X8)  =  64  sq.  in. 
a  =  24 
Siibstitutincr  in  formula,  -„- (B  +  b -f  4m 
we    have,  -^o*  (32  +  0  +  64)  =  384    cubic 
inches,  volume. 


13.  Required  the  contents 
of  a  cylindrical  ring,  the 
thickness  of  which  is  2 
inches,  and  the  inner  diame- 
ter  12   inches. 

Operation.  —  12  +  2  +  2  =  16 
inches,  diam.  of  outer  circle. 

16X3.1416  =  50.2656  in.  eir.  outer  circle. 
12  X  3.1416  =  37_^992      "       inner      "      - 

2;87.9648  [other  two. 

43.9824      "■      of  a  circle  half  the  sum  of  the 

Conceive  the  ring  to  be  cut  and  then  made  straight. 

It    is  now  an  equal  solid  in  the  form  of   a  cylinder,  any 

cross  section  of   the    ring   being  equal    to    either   base  or 

middle  section  of  the  cylinder,  and  the  altitude  of  the  equal 

solid  in  the  cylindrical  form  being  equal  to  one -half  the  sum 

of  the  inner  and  outer  circumferences  of  the  ring.    Hence, 

B=      12X3.1416   =   3.1416  sq.  in.  (Art.  60,  R.  Ill  c) 

b=      1-X3.1416   =   3.1416       *'  "  " 

4111  =  4(1^X3.1416)  =12.5664       ''  "  '' 

a  =  43.9824 
Substituting  in  formula,  f  (B  -|-  b  -|~  4m) 
we  have,  ^^-^--'-i  (3.1416  +  3.1416 +12. 5664)=138. 1751+ 

cubic  inches,  volume  of  the  cylinder. 
The  volume  of  the   cylinder  is  equal  to  the  volume  of 
the  cylindrical  ring  ;  hence,  the  volume  of  the  cylinirieal 
ring  equals  138.1751+  cubic  inches. 

C 


34 


ME^' a  U  RATION 


P\] 


14.  A  certain  monument  has  the 
form  of  the  frustum  of  a  square 
pyramid  ;  how  many  cubic  feet  of 
material  does  it  contain  if  it  is  240 
feet  high,  the  lower  base  3G  feet 
square,  and  the  upper  base  24  feet 
square,  provided  that  through  the 
center,  from  top  to  bottom,  there  is 
an  opening  in  the  form  of  a  frustum 
of  a  cone  whose  lower  diameter  is 
20  feet,  and  upper  diameter  16  feet  ? 


.30N 


-18-t 


Operation. — Frustum  of  pyramid  : 

B=      36X3G  =1,296  sq.  ft. 
b=      24X  24  =    576      " 
4m  =  4(30X30)=3,600      " 

Substituting    in    formula,    |  (B  -f-  b  +  4m) 

we  have,  ^  (1,296  +  576  +  3,600)=  218,880 

cubic  feet  volume  if  there  were  no  opening. 


-20- 


Frustum  of  cone  : 

20-f-2  =  10  R  of  B. 
16-4-2=    8  R  of  b. 
18 -f- 2=    9  R  of  M. 
B  =  10^  X  3.1416  =  314.16  sq.  ft. 
b=    8- X  3.1416  =  201.0624    " 
4m  =  4  (9-  X  3.1416)  =  1,017.8784  sq.  ft. 

Substituting  in  formula,  ^  (B-|-b  +  4m) 
we  have,  ^  (314.16  +  201.0624  + 
1,017.8784)  =  61,324.032  cubic  feet  in 
frustum  of  cone  to  be  deducted.  Hence, 
218,880  —  61,324.032  =  157,554.968 
cubic  feet  of  material. 


ILL UHTltATl  \  K    rUOBLKMS 


15.  Mr.  Brown  contracted 
to  build  an  embankment  for 
a  railroad  up  a  sliglitly 
inclined  plane,  the  bed  for 
the  track  to  be  level.  The 
embankment  was  to  be  1  mile 
long,  and  the  width  of  the 
bed  for  the  ties  12  feet  wide  ; 
the  perpendicular  measure- 
ment of  the  higher  end  was 
to  be  20  feet,  and  at  the  lower 
end  16  feet ;  the  width  of 
the  embankment  at  the  higher 
end  at  the  bottom  was  to  be 
3G  feet,  and  at  the  lower 
end  24  feet ;  what  was  the 
cost  at  20  cents  per  cubic 
yard  ? 

Operation.— ADEG,  IJKL,  and 
STUV,  which  are  B,  b,  and  M,  are 
all  trapezoids.     Hence, 

B=      (36+12^-2)20  =    480  sq.ft. 
b=      (24+12 -J- 2)10  =288     " 
4m  =  45  (30+12 -^2)18i=l, 512     " 
a  =  5,280  ft. 

Substituting  in  formula, 4(B+  b  +4m) 
we  have,  H"  (4S0  +  2S8  +  1,512)  = 
2,000,400  cubic  feet.  2,000,400 
-^-  27  =  74,311^  cubic  yards, 
74,3lU  X  20  cents  =  $14,802.22^ 
cost. 


^^\ 


m 


^*"4':-^ 


"^h^i^ 


en: 


36 


Mt\\\ii:}L-irwy 


16.  A  bridge -pier  of  solid  masonry  has  a  rectangular 
base  10  by  40  feet ;  the  top  of  the  pier  is  a  rectangle  6 
by  28  feet,  and  the  perpendicular  distance  between  the 
two  bases  is  18  feet ;  how  many  perches  of  material  in 
the  pier  ? 

Operation.— 

40 +  28h-  2  =  34  ft.  side  of  middle  section. 
10+    6-f-2=    8  ft.  end    "         "  " 

B  =       40  X  10  =  400  sq.  ft. 

b=       28  X    6=  168 
4m  =  4  (34  X  8)  =1,088        " 
a  =  18 

Substituting  in  formula,  |.  (B  +  b  -(-  4m) 
we  have,^o^(400  +  168  +  1,088)=  4,908  cubic  feet. 
4,968 -f- 24.75  =  200.72+  perches,  Ans. 


17.  A  saddler  uses  a  block  for  mak- 
ing horse  collars  shaped  like  a  cylin- 
droid.  Find  the  volume  of  such  a 
block  whose  lower  base  is  an  ellipse 
having  a  transverse  axis  8  inches  long, 
conjugate  axis  4  inches  long,  and  whose 
upper  base  is  an  ellipse,  the  trans- 
verse axis  being  4  inches  long,  the  con- 
jugate axis  2  inches  long,  if  the  altitude 
of  the  block  is  30  inches. 


rOLUMES    OF    PRISMS    AND    CYLINDERS  37 

Operation. — 

B=       4X2    X  3.1416    =25.1328  sq.  in.  (Art.  61.  R) 

b=       2X1    X  3.1416    =    6.2832       " 
4m  =  4  (3  X  HX  3.1416)  =56.5488       " 

a  =  30 
Substituting  in  formula,  |(B  +  b  +  4ra) 
we  have,  ^(25. 1328 +  6.2832 +  56.5488)=  439.824  cubic  inches,  Ans. 

{8  +  4-i-2  =  6  transverse  axis  of  M. 
4  +  2-^2  =  3  conjugate       "         " 


VOLUMES   OF   PRISMS   AND   CYLINDERS 

85.  Rule  I.  To  find  the  volume  of  a  prism  or  cylin- 
der :    MuUlphj  ilie  area  of  the  base  by  the  altitude. 

Rule  II.  Multiply  the  sum  of  both  end  areas  and  four 
times  the  area  of  a  section  half-tuay  between  them,  by  one- 
sixth  the  altitude.     KB  +  b  +  4m) . 

PROBLEMS 

In  a  prism  or  cylinder,  given  : 

1.  Side  of  square  base  4  feet,  altitude  30  feet,  to  find 
the  volume.  Ans.  480  cu.  ft. 

2.  Side  of  square  base  3i  feet,  altitude  8  feet,  to  find 
the  volume.  Ans.  98  cu.  ft. 

3.  Side  of  regular  hexagonal  base  8  inches,  altitude  10 
feet,  to  find  the  volume.  Ans.  11.546+  cu.  ft. 

4.  Sides  of  a  triangular  base  12,  15,  and  24  feet,  alti- 
tude 20  feet,  to  find  the  volume.      Ans.  1,472.676+  cu.  ft. 

5.  Diameter  of  the  base  4  feet,  altitude  30  feet,  to  find 
the  volume.  Ans.  376.992  cu.  ft. 

6.  Diameter  of  the  base  6  inches,  altitude  5  feet,  to 
find  the  volume.  Ans.  1,696.464  cu.  in. 


38  MENSURATION 

7.  Radins  of  the  base  6  feet,  altitude  20  feet,  tu  find 
the  vohime.  Ans.  2,261.952  cu.  ft. 

8.  Radius  of  the  base  10  feet,  altitude  30  feet,  to  find 
the  volume.  Ans.  9,424.8  cu.  ft. 

9.  Circumference  of  the  base  31.416  feet,   altitude  40 
feet,  to  find  the  volume.  Ans.  3,141.6  cu.  ft. 

10.  Circumference  of  the  base  80  feet,  altitude  50  feet, 
to  find  the  volume.  Ans.  25,464+  cu.  ft. 

11.  Find  the  cost  of  a  stick  of  timber  20  inches  square 
and  30  feet  long,  at  25  cents  a  cubic  foot.      Ans.  $20.83i. 

12.  Find  the  number  of  cubic  feet  in  a  log  30  feet 
long  and  24  inches  in  diameter  at  each  end. 

Ans.  94.248  cu.  ft. 

13.  Find  the  capacity  in  gallons  of  a  cylindrical  cis- 
tern measuring  8  feet  across  and  20  feet  deep. 

Ans.  7,520.256  gals. 

14.  Find  the  value  of  a  piece  of  moulding,  the  end  of 
which  is  a  triangle  whose  sides  are  respectively  3,  4,  and 
5  inches,  and  length  24  feet,  at  50  cents  per  cubic  foot. 

Ans.  50  cts. 

15.  What  is  the  comparative  weight  of  two  poles  of 
the  same  material,  one  being  3  inches  in  diameter  and  10 
feet  long,  and  the  other  6  inches  in  diameter  and  20  feet 
long?  Ans.  latter  =8  times  former. 

16.  Find  the  volume  of  a  prism  whose  base  contains 
8i  square  feet,  and  the  square  of  whose  height  equals  six 
times  the  number  of  square  feet  in  the  base. 

Ans.  57i  cu.  ft. 

17.  A  lime  kiln  in  the  form  of  a  prism  is  12  feet 
square  aiid  24  feet  high  ;  through  the  center  of  this,  from 
top  to  bottom,  there  is  a  cylindrical  opening  6  feet  in 
diameter;  required  the  number  of  perches  of  masonry  in 
the  kilu.  Ans.  112  218+P. 


PIBAMIDS    AND    CONES  39 

PYRAMIDS   AND   CONES 

DEFIWITIOWS 

86.  A  Pyramid  is  a  solid  having  for  its  base  a  poly- 
gon and  for  its  sides  plane  triangles,  all  terminating  at 
one  point,  called  the  vertex. 

When  the  base  is  regular  the  pj-ramid  is  regular ; 
otherwise  irregular. 

87.  A  Cone  is  a  solid  whose  base  is  a  circle,  and 
whose  lateral  curved  surface  tapers  uniformly  to  a  point, 
called  the  vertex. 

88.  The  Altitude  of  a  pyramid  or  of  a  cone  is  the 
perpendicular  distance  from  the  vertex  to  the  plane  of  the 
base. 

89.  The  Axis  of  a  pyramid  or  of  a  cone  is  the  straight 
line  that  joins  the  vertex  with  the  center  of  the  base. 

90.  A  Rig'ht  Pyramid  or  a  Rig-ht  Cone  is  one  whose 
axis  is  perpendicular  to  the  base  of  the  pyramid  or  cone. 
The  base  of  the  pyramid  is  a  regular  polygon. 

91.  An  Oblique  Pyramid  or  an  Oblique  Cone  is  one 
whose  axis  is  oblique  to  the  base  of  the  pyramid  or  cone. 

92.  The  Slant  Heig-ht  of  a  pyramid  is  the  perpen- 
dicular distance  from  its  vertex  to  either  of  the  sides  of 
the  base. 

93.  The  Slant  Height  of  a  cone  is  a  straight  line 
drawn  from  the  vertex  to  the  circumference  of  the  base. 

94.  A  Rig-ht   Cross  Section  of  a  pyramid   or  of  a 

cone  is  a  section  made  by  a  plane  perpendicular  to  the 
axis. 


40 


MENSURATION 


95.  A  Frustum  of  a  pyramid  or  of  a  cone  is  that  part 
which  remains  after  cutting  off  the  top  by  a  plane  paral- 
lel to  the  base. 

96.  The  Altitude  of  a  Frustum  of  a  pyramid  or  of 
a  cone  is  the  perpendicular  distance  between  the  bases. 

97.  The  Slant  Heigrht  of  the  Frustum  of  a  right 
pyramid  or  of  a  right  cone  is  the  shortest  distance 
between  the  perimeters  of  the  bases. 

98.  A  pyramid  is  triangular,  quadrangular,  pentago- 
nal, hexagonal,  etc.,  according  as  its  base  is  a  triangle, 
quadrilateral,  pentagon,  hexagon,  and  so  on. 


Riglit  Pyramid 

(Triangular) 

(.FUJ-  1^ 


Eight  Pyramid 

(Hexagonal) 

{Fig.  2) 


Oblique  Pyramid 
(Quadrangular) 
{Fig.  3) 


Eight  Cone 
{Fig.  4) 


In  Fig.  4,  if  the  right-angled  triangle  MON  be 
revolved  around  MO  as  an  axis,  it  will  generate  the  cone 
NQP— M. 

In  Fig.  7,  if  the  trapezoid  MNOP  be  revolved  around 
MO  as  an  axis,  it  Avill  generate  the  frustum  of  the  cone 
QRP— N. 

In  Figs.  1,  2,  3,  4,  5,  6,  and  7,  the  line  representing 
the  altitude  is  marked  a. 

In  Figs.  1,  2,  4,  6,  and  7,  the  line  representing  the 
slant  hei'^ht  is  marked  b. 


rVUAMlDS    AM)    (UM'IS 


41 


All  the  lateral  faces  of  a  ri^lit  pyramid  are  equal  tri- 
angles ;  the  sum  of  tlie  bases  of  these  triangles  is  equal 
to  the  perimeter  of  the  base  of  the  right  pyramid. 

The  altitude  of  each  lateral  triangle  is  equal  to  the 
slant  height  of  the  pyramid  ;  moreover,  a  cone  may  be 
regarded  as  a  pyramid  of  an  infinite  number  of  sides. 
The  reason  for  the   following  rule  is,  therefore,  obvious  : 

Rule  I.  To  find  the  conrex  surface  of  a  right  pyra- 
mid or  of  a  right  cone  :  Multiply  the  perimeter  ov  circum- 
ference  of  the  base  hy  one-half  the  slant  heUjht. 

To  find  the  entire  surface,  add  to  the  convex  surface 
the  area  of  the  base.  The  lateral  surface  of  any  })yra- 
mid  may  be  found  by  adding  together  the  areas  of  the 
lateral  faces. 


Oblique  Coue 
(Fig.  5) 


rrubtiiiu  of  Right 
Pyramid 


r 

0 

/ 

y-^ 

1 

ru; 

-llMll 

of 

i\ 

:•?;» 

ll  Co 

lie 

c^ 

lij.  < 

.1 

99.  Each  lateral  face  of  the  frustum  of  a  right  pyra- 
mid is  a  trapezoid.  The  slant  height  of  the  frustum  is 
equal  to  the  perpendicular  distance  between  the  parallel 
sides  of  the  trapezoid.  The  perimeter  of  the  lower  base 
of  the  frustum  is  equal  to  the  sum  of  the  lower  bases 
of  the  trapezoids  forming  the  convex  surface,  and  the 
perimeter  of  the  upper  base  of  the  frustum  is  equal 
to  the  sum  of  the  upper  bases  of  the  same  tra]>ezoids. 
Now,  the    frustum   of    a    cone    may    be  regarded   as  the 


42  MENS  URA  Tl  ON 

frustum  of  a  pyramid  of  an  iiitiuite  iium])er  of  sides. 
The  rule  for  finding  the  area  of  a  trapezoid  is,  therefore, 
applicable. 

Rule  II.  To  find  the  convex  surface  of  the  frustum 
of  a  rig-ht  pj'ramid  or  cone  :  Multiply  the  sum  of  the  perim- 
eters or  circumferences  of  the  two  bases  hij  one -half  the 
slant  height. 

For  the  entire  surface,  add  to  the  convex  surface  the 
area  of  both  ends  or  bases. 

Rule  III.  To  find  the  volume  of  a  pj'ramid,  cone, 
frustum  of  a  pyramid,  or  frustum  of  a  cone  :  Multiplf/ 
the  sum  of  hoth  end  areas  and  four  times  the  area 
of  a  section  half  ivay  l)etween  them,  by  one- sixth  the 
altitude.     -^  (B  +  b  +  4m) ,  or. 

To  find  the  volume  of  a  pyramid  or  cone  :  Multiply 
the  area  of  the  base  by  one-third,  the  altitude. 

To  find  the  volume  of  the  frustum  of  a  pyramid  or  of  a 
cone  :  To  the  sum  of  the  areas  of  both  bases  add  the 
square  root  of  their  product,  and  multiply  this  sum  by 
one -third  the  altitude. 


SURFACES  OF  PYRAMIDS  AND  CONES 
AND  THEIR  FRUSTUMS 

PROBLEMS 

1.  What  is  the  lateral  surface  of  a  quadrangular  pyra- 
mid, the  sides  of  whose  base  are  each  5  feet  and  slant 
height  30  feet?  Ans.  300  sq.  ft. 

2.  What  is  the  entire  surface  of  the  pyramid  mentioned 
in  Problem  1  ?  Ans.  325  sq.  ft. 


PYRAMIDS    AXl)    COXES  43 

3.  Wliat  is  the  lateral  surface  of  a  hexagonal  pyramid, 
the  sides  of  whose  base  are  each  8  feet  and  slant  height 
40  feet?  Ans.  960  sq.  ft. 

4.  Find  the  cost  of  painting  an  octagonal  church -spire 
at  22  cents  per  square  yard,  the  sides  of  whose  base  are 
each  4  feet  and  slant  height  60  feet.  Ans.  $23.46i. 

5.  Find  the  convex  surface  of  a  cone,  the  diameter  of 
whose  base  is  10  feet  and  slant  height  20  feet. 

Ans  314.16  sq.  ft. 

6.  Find  the  entire  surface  of  the  cone  mentioned  in 
Problem  5.  Ans.  392.7  sq.  ft. 

7.  How  many  square  yards  of  canvas  will  it  take  to 
make  a  conical  tent  whose  center -pole  is  8  feet  high  and 
the  diameter  of  the  base  12  feet!      Ans.  20.944  sq.  yd. 

8.  Required  the  lateral  surface  of  the  frustum  of  a 
square  pyramid  whose  slant  height  is  40  feet,  the  side  of 
the  lower  base  20  feet,  and  upper  base  16  feet. 

Ans.  2,880  sq.  ft. 

9.  AVhat  is  the  entire  surface  of  the  frustum  men- 
tioned in  Problem  8?  Ans.  3,536  sq.  ft. 

10.  What  is  the  lateral  surface  of  the  frustum  of  a 
hexagonal  pyramid  whose  ^lant  height  is  20  feet,  the  sides 
of  the  lower  base  each  6  feet,  and  upper  base  4  feet  ? 

Ans.  600  sq.  ft. 

11.  Find  the  convex  surface  of  the  frustum  of  a  cone 
whose  slant  height  is  30  feet,  the  circumference  of  the 
lower  base  40  feet,  and  of  the  upper  base  30  feet. 

Ans.  1,050  sq.  ft. 

12.  Find  the  entire  surface  of  the  frustum  mentioned 
in  Problem  11.  Ans.  1,248.943+  sq.  ft. 

13.  What  is  the  convex  surface  of  the  frustum  of  a 
square  pyramid  whose  slant  height  is  10  inches,  the  side 
of  the  upper  base  being  18  inches  and  of  the  lower  base 
20  inches?  Ans.  760  sq.  in. 


44  MENSURATION 

VOLUMES   OF    PYRAMIDS  AND   CONES 

PROBLEMS 

lu  a  pyramid,  given  : 

1.  The  side  of  a  square  base  4  feet,  altitude  30  feet, 
to  find  the  volume.  Ans.  160  cu.  ft. 

2.  The  side  of  a  square  base  8  inches,  altitude  21  feet, 
to  find  the  volume.  Ans.  3  cu.  ft.,  192  cu.  in. 

3.  The  area  of  the  square  base  16  square  j^ards,  alti- 
tude 30  yards,  to  find  the  volume.  Ans.  160  cu.  yd. 

4.  The  sides  of  a  rectangular  base  60  X  40  feet,  alti- 
tude 90  feet,  to  find  the  volume.  Ans.  72,000  cu.  ft. 

5.  Find  the  volume  of  a  square  pyramid  whose  alti- 
tude is  60  feet,  and  the  side  of  whose  base  is  9  feet. 

Ans.  1,620  cu.  ft. 

6.  The  base  of  a  right  triangular  pyramid  is  an 
equilateral  triangle,  each  side  of  which  is  12  feet  and  the 
altitude  18  feet  ;   find  the  cubic  contents. 

Ans.  374.1228+  cu.  ft. 

7.  The  altitude  of  a  pyramid  is  8  feet,  and  its  base  is 
a  rectangle  3  feet  by  2  ;   find  the  volume.    Ans.  16  cu.  ft. 

8.  A  certain  pyramid  has  an  altitude  equal  to  twice 
the  diagonal  of  its  base,  which  is  a  rectangle  60  by  80 
feet;   find  its  volume.  Ans.  320,000  cu.  ft. 

9.  Find  the  volume  of  a  pyramid  whose  base  is  a  rec- 
tangle 40  feet  by  30,  and  the  length  of  the  edges  which 
meet  at  the  vertex  65  feet.  Ans.  24,000  cu.  ft. 

10.  A  granite  monument  consists  of  a  pedestal  24 
inches  square  and  4  feet  high,  on  which  stands  a  pyramid 
18  inches  squai-e  and  9  feet  high;  what  did  it  cost  at 
$15  per  cubic  foot  ?  Ans.  $341.25. 


In  a  cone,  given  : 

11.  The  diameter  of  the  base  6  feet,  altitude  18  feet, 
to  find  the  volume.  Ans.  1G9.6464  cu.  ft. 

12.  The  diameter  of  the  base  2  feet,  altitude  12  feet, 
to  find  the  volume.  Ans.  12.5G64  cu.  ft. 

13.  The  radius  of  the  base  10  inches,  altitude  15 
inches,  to  find  the  volume.  Ans.  1,570.8  cu.  in. 

14.  The  circumference  of  the  base  62.832  feet,  altitude 
30  feet,  to  find  the  volume.  Ans.  3,141.6  cu.  ft. 

15.  Find  the  solid  contents  of  a  cone  whose  altitude 
is  48  feet  and  the  diameter  of  whose  base  is  5  feet. 

Ans.  314.16  cu.  ft. 

16.  How  many  tons  of  hay  in  a  conical  stack  30  feet 
high  and  40  feet  across  the  base,  there  being  500  cubic 
feet  to  the  ton?  Ans.  25.1328  tons. 

17.  Mr.  Johnston's  crop  of  wheat  piled  on  his  barn 
floor  formed  a  heap  in  the  form  of  a  cone  10  feet  high 
and  62.832  feet  in  circumference  ;  what  was  it  worth  at 
$1.25  per  bushel?  Ans.  $1,051,865. 

18.  Mr.  Brown's  granary  sprang  a  leak  2  feet  above 
the  level  of  his  barn  floor  ;  the  clover  seed  that  ran  out 
of  this  opening  formed  a  heap  in  the  shape  of  a  half -cone 
2  feet  high,  and  extended  2  feet  at  the  bottom  from  the 
perpendicular  partition  between  the  barn  floor  and  the 
granary;   how  many  bushels  leaked  out?  Ans.  3.3659+ bu. 

19.  The  circumference  of  the  base  of  a  cone  is  15.708 
feet,  and  the  altitude  is  equal  to  the  diameter  of  the  base  ; 
find  its  volume.  Ans.  32.725  cu.  ft. 

20.  Compare  the  volumes  of  a  cone  and  a  cylinder,  the 
altitude  of  each  being  12  feet  and  the  radius  of  the  base 
of  each  4  feet.  Ans.  Cone  =  i  vol.  eyl. 


4G  MKXSURATIOX 

VOLUMES   OF  FRUSTUMS  OF  PYRAMIDS  AND   CONES 

PROBLEMS 

Given,  in  the  frustum  of  a  pyramid  : 

1.  Upper  side  of  square  base,  10  feet;  lower  side, 
12  feet ;   altitude,  20  feet ;   to  find  the  volume. 

Ans.  2,426f  cu.  ft. 

2.  Upper  sides  of  rectangular  base,  12  and  IG  feet ; 
lower  sides,  18  and  24  feet ;  altitude,  30  feet ;  to  find 
the  volume.  Ans.  9,120  cu.  ft. 

3.  Upper  sides  of  triangular  base,  each  6  feet ;  lower 
sides  of  triangular  base,  each  8  feet;  altitude,  18  feet; 
to  find  the  volume.  Ans.  384.50+  cu.  ft. 

4.  Find  the  volume  of  the  frustum  of  a  square  pja^a- 
mid  whose  altitude  is  30  feet,  each  side  of  the  lower 
base  24  feet,  and  of  the  upper  base  18  feet. 

Ans.  13,320  cu.  ft. 

5.  Find  the  volume  in  cubic  feet  of  a  stick  of  timber 
36  feet  long,  the  larger  end  being  18  inches  square  and 
the  smaller  end  15  inches  square.  Ans.  68t  cu.  ft. 

6.  How  many  cubic  feet  in  the  upright  of  a  crane 
30  feet  high,  the  lower  end  being  2  feet  square  and  the 
upper  end  18  inches  square  ?  Ans.  922  cu.  ft. 

7.  The  lower  base  of  the  frustum  of  a  pyramid  is  a 
rectangle  20  by  40  inches  ;  the  upper  base,  a  rectangle 
15  by  30  inches,  and  the  altitude  12  feet ;  what  is  the 
volume  I  Ans.  51  cu.  ft.  672  cu.  in. 

8.  The  smaller  base  of  the  frustum  of  a  pyramid  is 
a  rectangle  3  by  4  feet ;  the  longer  side  of  the  lower 
base  is  equal  to  the  diagonal  of  the  upper  base,  and  the 
shorter  side  of  the  lower  base  bears  the  same  ratio  to 
the   longer  side    as    the    shorter  side  of    the    upper  "base 


}  DLrvKs  or  rinsTiwfs  or  ryu.t.yins  jxn  ro.v/;\   47 

hears  to  its  longer  side  ;  the  altitude  is  equal  to  the 
sum  of  the  perimeters  of  both  bases.  Required  the 
volume.  Ans.  480.375  cu.  ft. 

9.  What  are  the  solid  contents  of  the  frustum  of  a 
cone  Avhose  upper  l)ase  is  10  feet  in  diameter,  lower  base 
16  feet,  and  altitude  18  feet?  Ans.  2,431.5984  cu.  ft. 

10.  Find  the  number  of  eubie  feet  in  a  log  42  feet 
long,  the  diameter  of  the  lai-ger  end  being  3  feet  and 
of  the  smaller  end  2   feet.  Ans.  208.9164  eu.  ft. 

11.  A  telephone  pole  48  feet  long  is  5  feet  in  cir- 
cumference at  the  bottom  and  3  feet  at  the  top;  required 
the  contents.  Ans.  62.387+  cu.  ft. 

12.  Find  the  capacity  in  gallons  of  a  pail  18  inches 
cieep,  measuring  14  inches  across  the  top  and  12  inches 
across  the  bottom.  Ans.  10.363+  gal. 

13.  A  tub  2  feet  deep  is  26  inches  in  diameter  at  the 
bottom  and  30  inches  at  the  top  ;  how  many  pails  of 
water  the  size  of  the  one  described  in  Problem  12  will 
be  required  to  fill  the  tub!  Ans.  6.1+  pails. 

14.  Mr.  Long's  cistern  is  7  feet  deep,  the  bottom 
being  6  feet  in  diameter  and  the  top  4  feet ;  how  many 
barrels  of  water  will  it  hold  f  Ans.  33.07+  bbl. 

15.  How  many  quarts  of  wine  will  be  required  to  fill 
2  dozen  tumblers,  each  2  inches  across  the  bottom,  3  inches 
across  the  top,  and  4  inches  deep?  Ans.  8.26+  qt. 

16.  A  butcher's  pickle -stand  is  5  feet  deep,  6  feet  in 
diameter  at  tlie  bottom,  and  4  feet  at  the  top  ;  how^  many 
gallons  will  it  contain  '?  Ans.  744.192  gal. 

17.  A  farmer  had  a  conical  stack  of  hay  314.16  feet 
in  circumference  at  the  bottom  and  36  feet  high.  A 
storm  blew  away  the  top  so  that  the  remainder  was  fo\md 


4S 


Mi:XS(JiATI()X 


to  be  only  18  feet  high ;    how  many  cubic  feet  remained 
in  the  stack "?     How  many  cubic  feet  blew  away  ? 

Ans. Remained,  82,467  cu.  ft.;   blew  away,  11,781  cu.  ft. 

18.  At  $15  per  cubic  foot,  find  the  difference  in  the 
cost  of  two  granite  monuments,  one  of  which  is  the  frus- 
tum of  a  square  pyramid  3  feet  square  at  the  top,  4  feet 
square  at  the  bottom,  and  30  feet  high  ;  the  other  is  the 
frustum  of  a  cone  30  feet  high,  the  diameter  of  the  upper 
base  being  3  feet  and  of  the  lower  base  4  feet. 

Ans.  $1,191.03. 

THE   SPHERE 


DEFimXIOWS 

100.  A  Sphere  is  a  solid  bounded  by  a  uniformly 
curved  surface,  evei-y  point  of  which  is  equally  distant 
from    a   point    within,   called    the 

center.  It  may  be  generated  by 
the  revolution  of  a  semi- circle 
about  its  diameter  as  an  axis. 

101.  A  Diameter  of  a  sphere 
is  a  straight  line  passing  through 
the  center  of  the  sphere,  and  termi- 
nated at  V)oth  ends  bv  the  surface. 


^.Ktcua 

If^^e 

/m 

'■-HX\ 

?^rilJn^ 

e     t  el  V    ^"^1 

\\V" 

^W7 

\\\  \ 

1  f/y 

\^ 

^>^ 

Sphere 
FigX 


102.  A  Radius  of  a  sphere  is 
a  straight  line  drawn  from  the 
center  to  any  point  in  the  surface.     It  is  half  the  diameter. 


103.  A  Zone  is  the  portion  of  the  surface  of  a  sphere 
included  between  two  parallel  planes. 

104.  A  Spherical  Segment  is  the  solid  portion  of  a 
sphere  included  between  two  parallel  planes. 


THE  s rut: HE  49 

105.  A  Spherical  Sector  is  a  volume  generated  by 
the  revolution  of  a  circular  sector  al)out  the  diameter. 

106.  A  Great  Circle  is  a  section  made  by  any  plane 
passing  through   the  center  of  a   sphere. 

Rule  I.  To  find  the  surface  of  a  sphere  :  Multiply  the 
circumferenci  of  the  sphere  hy  the  diameter.     Or,  S  =  7rD^. 

Rule  II.  Multiply  the  square  of  the  radius  by  four  times 
3.1416.     Or,  8  =  47^R^ 

Rule  III.  To  find  the  volume  of  a  sphere  :  Multiply 
the  sum  of  both  end  areas  and  four  times  the  area 
of  a  section  half  way  between  them,  by  one- sixth  the  alti- 
tude.    |(B-f-b+4m) 

Rule  IV.  Multiply  the  surface  by  one- sixth  the  diam- 
eter, or  one -third  the  radius. 

Rule  V.     Multiply  the  cube  of  the  diameter  by  one-sixth 

of  3.1416,  or  by   .5236.     Or,  V  = 'L^. 

107.  The  spliere  may  be  regarded  as  a  polyedron 
consisting  of  an  infinite  number  of  pyramids  whose  com- 
bined bases  form  the  surface  of  the  sphere  and  whose 
equal  altitudes  are  equal  to  the  radius  of  the  sphere. 
Since  the  volume  of  each  pyramid  is  equal  to  its  base 
multiplied  by  one -third  of  its  altitude,  and  since  the 
whole  is  equal  to  the  sum  of  all  its  parts,  Rule  IV  is 
readily  derived. 

The  cylinder,  the  cone,  and  the  sphere  are  the  three 
round  bodies  of  Geometry.  There  is  a  fixed  relation 
existing  between  these  three  solids,  both  as  to  their  sur- 
faces and  their  volumes.  See  Problems  22  and  23,  Art. 
il4. 


.')( )  MEXS  URA  TI  ON 

SURFACES    OF  THE    SPHERE 

PROBLEMS 

111  a  sphere,  given  : 

1.  The  diameter  8  inches,  to  find  the  surface. 

Ans.  201.06+  sq.  in. 

2.  The  diameter  18  inches,  to  find  the  surface. 

Alls.  1,017.9 —  sq.  in. 

3.  The  diameter  10  inches,  to  find  the  surface. 

Ans.  314.16  sq.  in. 

4.  The  radius  3  inches,  to  find  the  surface. 

Ans.  113.1 —  sq.  in. 

5.  The  radius  8  inclies,  to  find  the  surface. 

Ans.  804.25 —  sq.  in. 

6.  The  radius  10  inches,  to  find  the  surface. 

Ans.  1,256.6+  sq.  in. 

7.  The     circumference     62.832     inches,    to    find     the 
surface.  Ans.  1,256.6+  sq.  in. 

8.  The     circumference     15.708     inches,    to     find     the 
surface.  Ans.  78.54  sq.  in. 

9.  The     circumference    251.328    inches,    to    find    the 
surface.  Ans.  20,106.24  sq.  in. 

10.  How   many   square    inches  of  tissue  paper  will  be 

required   to  wrap  20  oranges  each  2  inches  in  diameter, 

provided  no  allowance  is  made  for  waste  ? 

Ans.  251.328  sq.  in. 

11 .  How  many  square  inches  of  pig  skin  in  a  spherical 
football  9  inches  in  diameter?  Ans.  254.47 —  sq.  in. 

12.  How  many  square  inches  of  bronze  will  it  take 
to  cover  the  three  balls  used  as  a  sign  of  a  pawn- 
broker, if  each  ball  is  4  inches  in  diameter  ? 

Ans.  150.796+  sq.  in. 


voiA.MEs   OF   THE   sriiu:!-:  51 

VOLUMES    OF  THE    SPHERE 

PROBLEMS 

In  a  sphere,  given  : 

1.  The  diameter  2  feet,  to  find  the  volume. 

Ans.  4.1888  en.  ft. 

2.  The  diameter  5  inches,  to  find  the  volume. 

Ans.  65.45  cu.  in. 

3.  The  radius  23  inches,  to  find  the  volume. 

Ans.  50,965.1296  cu.  in. 

4.  The  circumference  37.6992  feet,  to  find  the  volume. 

Ans.  904.7808  cu.  ft. 

5.  How  much  will  16  croquet  balls  each  4  inches  in 
diameter  weigh,  if  a  cubic  inch  of  the  wood  of  which 
they  are  made  weighs  i  ounce!       Ans.  16  lbs.  12.08+  oz. 

6.  If  the  croquet  balls  mentioned  in  Problem  5  were 
placed  in  a  rectangular  box  16  inches  long,  8  inches  deep, 
and  8  inches  wide,  how  many  quarts  of  sand  (dry  measure) 
would  be  required  to  fill  all  the  remaining  space  and  make 
the  box  level  full  f  Ans.  7.259+  qt. 

7.  Admiral  Dewey  at  the  battle  of  Manila  used  a  num- 
ber of  8 -inch  guns  ;  what  would  be  the  weight  of  a  solid 
spherical  ball  made  to  fit  these  guns,  if  a  cubic  foot  of 
iron  weighs  450  pounds?  Ans.  69.81251b. 

8.  A  sphere  is  20  inches  in  diameter  ;  what  is  the 
volume  of  an  inscribed  cube  f         Ans.  1,539.598+  cu.  in. 

9.  The  edge  of  a  cube  is  20  inches  ;  what  is  the  volume 
of  an  inscribed  sphere  ?  Ans.  4,188.8  cu.  in. 

10.  The  outside  diameter  of  a  hollow  globe  of  uniform 
thickness  is  16  inches  and  the  inside  diameter  14  inches  ; 
how  roany  cubic  inches  of  metal  in   the   shell  ? 

Ans.  707.9072  cu.  in. 


52  MEy.suiuriox 

11.  The  main  battery  of  the  battleship  Oregon  has 
four  13 -inch,  eight  8 -inch,  and  four  6 -inch  guns  ;  if  solid 
splierical  balls  were  used  instead  of  projectiles,  what 
weight  of  metal  would  be  required  to  load  these  once 
round,  there  being  450  pounds  of  the  metal  used  to  the 
cubic  foot?  Ans.  1,874.59+lb. 

12.  A  hollow  sphere  having  an  inside  diameter  of  12 
inches  was  filled  with  water  ;  if  a  solid  ball  3  inches  in 
diameter,  a  solid  cylinder  having  a  base  2  inches  in 
diameter  and  4  inches  high,  and  a  solid  square  pyramid 
whose  base  is  3  inches  on  a  side  and  altitude  6  inches, 
were  placed  inside  the  sphere  so  as  to  displace  part  of 
the  water,  how  many  cubic  inches  of  water  would 
remain!  Ans.  860.0772  cu.  in. 

THE    SPHEROID 


DEFINITIONS 

108.    A  Spheroid  is  a  volume  formed  by  the  revolu- 
tion of  an  ellipse  about  one  of   its  axes.     A  revolution 
about    the    longer    axis    forms    a 
prolate   spheroid ;     and    about   the 
shorter  axis,  an  oblate  spheroid. 

Rule  I.  To  find  the  volume  of 
a  spheroid :  Multiply  the  sum 
of  both  end  areas  and  four  times 
the  area  of  a  section  half  way  be- 
tween them,  by  one -sixth  the  alti- 
tude.    i(B  +  b  +  4m). 

Note. —  For  the  spheroid  each  end  area  is  Zero,  and  either  axis 
may  be  used  as  the  altitude,  according  as  the  area  of  the  middle 
section  is  found  for  the  prolate  or  oblate  spheroid. 

Rule  II.  Multiply  the  square  of  the  revolving  axis  hu 
the  fixed,  axis  and  this  product  by  .5236. 


Spheroid 

or 

Ellipsoid 

Fig.l 


VOLUMES    OF    THE    SPHEROID  53 

VOLUMES    OF  THE    SPHEROID 
PROBLEMS 

1.  What  is  the  vohime  of  a  spheroid  whose  longer 
diameter  is  24  iuches  and  shorter  diameter  20  inches  ? 

Ans.  5026.56  cu.  in. 

2.  Find  the  volume  of  a  stone  in  the  shape  of  a 
spheroid  whose  longer  diameter  is  4  feet  and  shorter 
diameter  2  feet.  Ans.  8.'J776  cu.  ft. 

o.  A  watermelon  is  shaped  like  a  spheroid ;  what 
are  its  contents,  if  its  longer  diameter  is  18  inches  and 
shorter  diameter  12  inches"?  Ans.  1,357.1712  cu.  in. 

4.  How  many  cubic  inches  of  air  in  a  spheroidal 
football  whose  longer  inside  diameter  is  14  inches  and 
shorter  diameter  9   inches?  Ans.  593.7624  cu.  in. 

5.  Which  has  the  greater  volume  and  how  much,  a 
sphere  12  inches  in  diameter,  or  a  spheroid  whose  longer 
axis   is  8  inches   and  shorter  axis  4  inches  f 

Ans.  The  sphere,  837.76  cu.  in.  greater. 

6.  Which  has  the  greater  volume  and  how  much,  a 
spheroid  whose  longer  diameter  is  30  inches  and  shorter 
diameter  20  inches,  or  the  frustum  of  a  square  pyramid 
whose  lower  base  is  10  inches  square,  upper  base  8 
inches  square,  and  whose  altitude  is  30  inches  ? 

Ans.  The  spheroid,  3843.2  cu.  in.  greater. 

7.  A  sphere  whose  radius  is  8  inches  and  a  sphei'oid 
having  a  longer  diameter  of  18  inches  and  shorter  diam- 
eter of  12  inches,  were  placed  in  a  tub  in  the  form  of  a 
frustum  of  a  cone  ;  the  lower  diameter  of  the  tub  was 
2  feet,  the  upper  diameter  26  inches,  and  depth  18 
inclies ;  how  many  gallons  of  water  would  have  been 
required  to  fill  the  remaining  space?       Ans.  23.11-f  gai. 


54 


MENSURATION 


CIRCULAR    RINGS 

DEFINITIONS 

109.    A  Circular  Ring"  is  formed  by  bending  a  cylin- 
der   or    a    bar  nut  11    the 


Circular  Rings 


two  ends  meet. 

It  will  be  observed  by 
examining  Figs.  1  and  3 
that  if   the    ring    is    cut 
at  any    place    and   made 
straight  the  same  volume 
will  be  changed  into  the 
form  of  a  cylinder.     The 
altitude    of   the   cylinder 
will   equal    one -half   the 
sum    of    the    inner    and 
outer  circumferences,  and    either 
base  of  the  cylinder  will  be  equal 
to  any  cross  section  of  the  ring. 
The    rules   for   finding   both    the 
surface  and  the  volume,  therefore, 
are  practically  the  same  as  those 
for    finding    the    convex    surface 
and  volume  of  a  cylinder. 

Rule  I.  To  find  the  surface  of  a  circular  ring  : 
Multiply  the  perimeter  of  a  cross  section  of  the  ring, 
or  its  girt,  by  one -half  the  sum  of  the  inner  and  outer 
circumferences. 

Rule  II.     To  find  the  volume  of  a  circular  rhig  : 
Multiply    both  end   areas   (conceiving   the   ring   to   he    pi^ 
made  straight)  and  four  times    the  area  of  a  section 
half   way    between    them,  by    one -sixth    the    altitude. 

|(B  +  b-j-4m). 

Rule  III.    Multiply  the  area  of  a  rrosH  sortion  of  the  ring 
by  one -half  the  sum  of  the  inm  r  and  outer  circumfprrnces. 


Fig.Z 


X 


SriiFACl'JS    OF    ClRl'CLAR    JilXGS  55 

SURFACES    OF   CIRCULAR    RINGS 

PROBLEMS 

1.  What  is  the  surface  of  a  pneumatic  bicycle  tire 
■whose  outside  diameter  is  28  inches  and  thickness 
li   inches?  Ans.  392.3186+  sq.  in. 

2.  A  hose  whose  outside  diameter  is  2  inches  was 
placed  in  the  form  of  a  ring- ;  the  outer  circumference 
measured  62.832  feet ;   required  the  surface  of  the  hose. 

Ans.  32.6246  +  sq.ft. 

3.  The  felloes  of  a  heavy  lumber  wagon  were  made 
2  inches  deep  and  4  inches  wide  ;  the  outside  diameter 
of  each  hind  wheel,  inside  the  tire,  was  4  feet  4  inches  ; 
how  many  square  f  ^et  of  surface  are  there  to  paint 
on  each  felloe  of  the  hind  wheels,  provided  the  four  sides 
are  painted  before  putting  on  the  tire,  and  no  allowance 
is  made  for  shaving  off  the  corners  on  the  inside  ? 

Ans.  13.09  sq.  ft. 

4.  A  semi -circular  white  oak  arch  of  uniform  thick- 
ness and  3  inches  in  diameter,  was  made  to  span  the 
main  street  of  a  town ;  the  distance  on  the  inside 
between  the  ends  of  the  poles  where  buried  in  the  ground 
was  30  feet ;  required  the  number  of  square  feet  of  bun- 
ting needed  to  cover  the  arch.  Ans.  37.3196+  sq.  ft. 

VOLUMES    OF   CIRCULAR    RINGS 

5.  What  is  the  volume  of  a  solid  iron  ring  of  the 
same  dimensions  as  the  pneumatic  bicycle  tire  in 
Problem    1  ?  Ans.  147.1194+  cu.  in. 

6.  Find  the  solid  contents  of  a  grape  vine  of  exactly 
the  same  shape  and  dimensions  as  the  hose  in  Problem  2. 

Ans.  1.3593+  cu.  ft. 


56 


MENSURATION 


7.  Find  the  number  of  cubic  inches  in  each  felloe 
described  in  Problem   3.  Ans.  1,256.64  cu.  in. 

8.  What  is  the  weight  of  the  white  oak  arcii  in  Prob- 
lem 4,  if  a  cubic  foot  of  wood  weighs  ol  pounds? 

Ans.  118.956+  lb. 

THE   WEDGE 

DEFHriTIOWS 

110.  A  Wedg'e  is  a  volume  of  five  sides,  having  a 
rectangular  base,  two  rectangular  or  trapezoidal  sides 
meeting  in  an  edge,  and  two  triangular  ends. 


Fig.\ 


Fi(/.2 


Wcdgt 


Fig:d 


FigA 


Rule  I.  To  find  the  surface  of  a  wedge  :  Find  the 
sum  of  the  areas  of  the  jive  faces. 

Rule  XL  To  find  the  volume  of  a  wedge  :  Midtiphj 
the  sum  of  the  two  end  areas  and  four  times  the  area  of 
a  section  half  way  between  them,  by  one-sixth  the  altitude. 
i(B  +  b  +  4m). 

In  the  wedge,  the  end  area  where  the  sides  meet  is  Zero. 


SURFACES  OF  THE  WEDGE 
PROBLEMS 

To  find  the  surface  of  a  wedge,  given  ; 
1.    Base  of  wedge,  a  rectangle  6x4  inches;   sides,  equal 
•ectangles ;    ends,  equal    isosceles    triangles ;     altitude    of 


SUIiFACliS    or    THE     WKDCE  57 

wedge,  12  inches;    altitude   of   rectaiio;le,   12.1655  inches. 
(Fig-.  2.)  Ans.  217.986  sq.  in. 

Note.  —  The  altitudes  of  the  wedgos  are  given  in  these  surface 
problems  ;  from  these  the  pupil  may  be  required  to  find  the  alti- 
tudes of  the  ends  and  sides,  when  time  will  permit. 

2.  Base  of  wedge,  a  square  4x4  inches;  sides,  equal 
trapezoids  4  inches  along  the  base  and  8  inches  along 
sharp  edge  ;  ends,  two  equal  isosceles  triangles  ;  altitude 
of  wedge,  6  inches  ;  altitude  of  each  trapezoid,  6.3246 
inches  ;   altitude  of  each  triangle,  6.3246  inches.    (Fig.  3.) 

Ans.  117.1936  sq.  in. 

3.  Base  of  wedge,  a  square  ^^Q>  inches  ;  sides,  2  equal 
trapezoids  whose  edges  next  the  base  are  each  6  inches 
and  sharp  edge  4  inches  ;  ends,  two  equal  isosceles  tri- 
angles ;  altitude  of  wedge,  8  inches  ;  altitude  of  trape- 
zoids, 8.544  inches  ;  altitude  of  triangles,  8.0623  inches. 
(Fig.  4.)  Ans.  169.8138  sq.  in. 

VOLUMES   OF   THE   WEDGE 

To  find  the  volume  of  a  wedge,  given  : 

4.  The  base  of  the  wedge,  a  rectangle  16  x  12  inches  ; 
the  sharp  edge,  16  inches  ;  triangular  ends  both  perpen- 
dicular to  the  base  of  the  wedge  ;  altitude  of  the  base,  21 
inches.     (Fig.  2.)  Ans.  2,016  cu.  in. 

5.  The  base  of  the  wedge,  a  square  12  x  12  inches;  the 
sharp  edge,  20  inches ;  the  altitude  of  the  wedge,  30 
inches.      (Fig.  3.)  Ans.  2,640  cu.  in. 

6.  The  base  of  the  wedge,  a  square  8x8  inches  ;  sharp 
edge,  6  inches  ;   altitude  of  wedge,  12  inches.      (Fig.  4.) 

Ans.  352  cu.  in. 

7.  A  plane  was  passed  from  the  upper  edge  of  a  cube, 
which  edge  is  6  inches,  to  the  lower  parallel  edge  on  the 
opposite  side  of  the  cube  :  i*e(iuired  the  volume  of  eitlier 
wedge  thus  made.      (Fig.  1.)  An.->.  108  cu.  in. 


58  MEX!SURAri(JX 

,     SIMILAR    SOLIDS 

DEFINITIONS 

111.  Similar  Solids  are  such  as  have  the  same 
form,  and  differ  from  each  other  only  in  size  ;    as,  cubes, 

i  spheres,  etc. 

112.  A  Dimension  of  a  solid  is  a  radius,  diameter, 
circumference,  height,  length,  breadth,  etc. 

113.  The  following  principles  of  similar  solids  are 
derived   from    geometry : 

Principle  I.  Similar  solids  are  to  each  other  as  the 
cubes  of  their  Wke  dimensions. 

Principle  II.  Lil-e  dimensions  of  similar  solids  are  to 
each  other  as  the  cube  roots  of  those  solids, 

PROBLEMS 

1.  If  the  volume  of  a  cube  4  inches  on  each  side  is 
64  cubic  inches,  what  is  the  volume  of  one  5  inches  on 
each  side  ? 

Operation. —  4^  :  5^  ::  (34  cu.  in.  :  X.        X  =  12.5  eii.  in.,  Ans. 

2.  How  many  baseballs  each  2  inches  in  diameter 
are  equal  in  volume  to  one  large  ball  1  foot  in 
diameter!  Ans.  21G. 

3.  If  a  conical  stack  of  wheat  in  the  sheaf  20  feet 
high  contains  400  bushels,  how  many  bushels  should  a 
similar  stack  30  feet  high  contain?  Ans.  1,350. 

4.  A  farmer  can  load  on  wagons  the  contents  of  a 
cubical  bin  10  feet  on  a  side  in  half  a  day  ;  how  long, 
at  the  same  rate,  will  it  take  him  to  load  the  contents 
of  a  cubical  bin  20  feet  on  a  side  ?  Ans.  4  days. 


SIMILAR    SOLIDS  59 

5.  If  a  colt  13^-  hands  high  weighs  1,000  pounds, 
how  much  should  he  weigh  when  he  becomes  a  horse 
15   hands  high?  Ans.  1,371.7+  lbs. 

G.  If  a  barrel  whose  bung -diameter  is  38  inches, 
when  filled  contains  G-1  gallons,  what  would  be  the  bung- 
diameter  of  a  similar  cask  that  holds  but  27  gallons  ? 

Ans.  28^  in. 

7.  If  the  volume  of  a  pyramid  whose  altitude  is  25 
feet  is  729  cubic  feet,  what  is  the  altitude  of  a  similar 
pyramid   whose   volume    is   9,261   cubic   feet? 

Ans.  58i  ft. 

8.  If  a  cylindrical  tank  2  feet  in  diameter  contains 
4  barrels  of  water,  what  is  the  capacity  of  a  similar 
tank  4  feet   in  diameter?  Ans.  32  bbl. 

9.  John  is  4  feet  tall  and  weighs  64  pounds ;  what 
should  be  his  height  when  he  weighs  216  pounds,  pro- 
vided he   develops   proportionately?  Ans.  6  ft. 

10.  If  it  requires  311  tons  of  ice  to  fill  a  cubical 
ice  house  whose  edge  is  10  feet,  how  many  tons  will  be 
required  to  fill  one  whose  edge  is  20  feet  ? 

Ans.  250  tons. 

11.  The  diameters  of  three  globes  are  respectively  i,  «, 
and  f  of  an  inch  ;  what  is  the  diameter  of  a  globe  equal 
to  the  volume  of  the  three?  Ans.  f  in. 

12.  The  patriotic  citizens  of  a  certain  city  wished  to 
raise  $8,000  for  the  purpose  of  erecting  a  monument  to 
the  heroes  Avho  went  down  with  the  "Maine"  in  Havana 
liarbor.  The  monument  was  to  be  the  frustum  of  a 
square  pyramid  50  feet  high.  Patriotism,  ran  so  high, 
however,  that  the  sum  of  $15,625  was  raised;  how  high 
a  monument  did  the  latter  sum  erect,  provided  the  orig- 
inal design  of  the  monument  was  carried  out? 

Ans.  62^  ft. 


GO  MENSURATION 

MISCELLANEOUS   PROBLEMS 

114.    Review. 

1.  Required  the  distance  between  an  upper  corner  and 
an  opposite  lower  corner  of  a  cliurch  96  feet  long,  72  feet 
wide,  and  25  feet  high.  Ans.  122.576+  ft. 

2.  How  many  square  yards  of  plastering  were  required 
to  cover  the  walls  and  ceiling  of  the  church  described  in 
the  preceding  problem,  deducting  80  square  yards  for 
openings?  Ans.  1,6213"  sq.  yds. 

3.  John's  bicycle  wheels  are  each  28  inches  in  diame- 
ter;  how  many  times  will  they  revolve  in  going  8,796.48 
inches  ?  Ans.  100  times. 

4.  A  gas  receiver  in  a  certain  city  is  a  cjdinder  40  feet 
in  diameter  ;  on  an  average  during  the  j'ear  each  morning 
it  is  6  feet  deeper  in  the  water  than  the  evening  before  ; 
how  many  cubic  feet  of  gas  are  consumed  each  night? 

Ans.  7,539.84  cu.  ft. 

5.  The  diameter  of  a  cylindrical  vessel  filled  with 
water  is  10  inches.  An  immersed  stone  displaces  2 
inches  of  the  depth  of  the  water.  How  many  cubic 
inches  are  there  in  the  stone?  Ans.  157.08  cu.  in. 

6.  A  teamster  hauled  from  a  mill  40  quadrangular 
posts,  each  6  by  8  inches  and  12  feet  long.  How  many 
triangular  posts,  the  base  of  each  of  which  is  a  triangle 
measuring  6,  8,  and  10  inches  on  a  side  respectively,  and 
the  same  length,  could  he  have  hauled,  provided  the 
loads  were   equal    in   weight  ?  Ans.  80  posts. 

7.  A  hall  60  by  120  feet  has  a  roof  in  the  form  of  a 
wedge.  Both  ends  are  equal  triangles  and  both  sides 
equal  trapezoids  ;    tlie  two  sides  meet  at  the  ridge,  which 


MiSCKLLAXEors    PROBIJIMS  61 

is  80  feet  long  and  30  feet  above  the  top  of  the  walls. 
Required  the  length  of  a  rafter  extending  from  a  corner 
of  the  building  to  the  ridge  f  Ans.  46.904+  ft. 

8.  Required  the  number  of  square   feet   in   the  entire 
roof  as  described  in  the  preceding  problem. 

Ans.  10,648.58-1- sq.ft." 

9.  How  many  cubic  feet  of  air  are  enclosed    by   the 
roof  and  attic  floor  of  the  hall  described  in  Problem  7  ? 

Ans.  96,000  cu.  ft. 

10.  In  cutting  the  grass  on  a  lawn  60  feet  wide  and 
containing  4,800  square  feet,  how  many  times  must  a 
lawn-mower  18  inches  wide  be  drawn  over  it  lengthwise 
so  that  the  whole  may  be  cut?  Ans.  40  times. 

11.  Mr.  Clark,  wishing  to  give  his  three  boys  prac- 
tical lessons  in  both  industry  and  foresight,  made  the 
following  proposition :  "  You  may  each  have  sufficient 
lumber  to  build  60  rods  of  fence,  with  which  you  may 
enclose  as  much  land  as  you  can  ;  all  that  you  raise  on 
the  land  thus  enclosed  ma}^  be  your  own  ;  I  should  like 
to  see  you  make  as  much  of  my  offer  as  you  possibly 
can."  The  oldest  enclosed  a  cirr^ular  piece  ;  the  second, 
a  square  piece,  and  the  youngest,  a  rectangular  piece 
twice  as  long  as  it  was  broad  ;  how  many  square  rods 
did  each  have  in  his  field  f 

Ans.  Oldest,  286.478+  sq.  rds.;  second,  225  sq.  rds.; 
youngest,  200  sq^  rds. 

12.  Find  the  length  of  a  hand-rail  for  a  flight  of 
stairs  having  20  steps,  each  step  ll  inches  high  and 
9    inches  wide.  Ans.  19.525+  ft. 

13.  In  Brandon  Park  there  is  a  regular  octagonal 
stand.  Surrounding  this  stand  there  is  a  porch  of  uni- 
form width.     The  perimeter  of  the  stand  is  6I3  feet;   the 


{\'2  MJ:Xsri!ATf()X 

perimeter  of  the  outer  edge  of  the  porch  is  94f  feet ;  the 
perpendicular  distance  between  the  stand  and  the  outer 
edg-e  of  the  porch,  or  the  width  of  the  porch,  is  5  feet  2 
inches.  How  many  square  feet  of  flooring  in  the  porch 
floor!  Ans.  403  sq.  ft. 

14.  The  frame  and  handle  of  a  corn -popper  is  made  of 
a  continuous  piece  of  wire.  The  handle  is  12  inches  long, 
and  the  frame  upon  which  the  wire  sieve  is  fastened  is  a 
rectangle  8  by  4  inches  ;  how  long  a  wire  was  required  to 
make  this  frame  and  handle,  making  no  allowance  for  the 
thickness  of  the  wire  ?  What  is  the  combined  length  of 
the  handle  and  frame?  Ans.  36  in.;   20  in. 

15.  A  lamp  having  a  paper  shade  is  placed  on  a  stand 
in  the  center  of  a  room  18  by  20  feet  ;  the  shade  casts  on 
the  ceiling  a  shadow  in  the  form  of  a  circular  ring  extend- 
ing the  entire  width  of  the  ceiling  and  uniformly  5  feet 
wide.  How  many  square  feet  of  the  ceiling  are  not 
shaded?  Ans.  155.796  sq.  ft. 

16.  Mr.  Jones  gave  each  of  his  two  sons  half  a  dollar 
to  spend.  The  younger  invested  his  in  5 -cent  tickets  for 
a  merry-go-round,  each  ticket  allowing  him  to  ride  on 
the  merry-go-round,  80  feet  in  circumference,  10  times. 
The  older  bought  a  railroad  ticket  at  3  cents  a  mile.  How 
many  miles  more  did  the  older  ride  than  the  younger  ? 

Ans.  15.15  mi. 

17.  The  width  of  Mr.  F's  buggy  from  center  of  tire  to 

center  of  tire  is  4  feet  9  inches  ;  the  diameter  of  the  fore 
wheel  is  43  inches  and  of  the  hind  w^heel  46  inches.  This 
buggy  is  drawn  round  a  circular  race -course  so  that  the 
outside  fore  wheel  leaves  a  track  exactly  one  mile  long. 
The  buggy  is  so  coupled  that  the  track  made  by  the  out- 
side hind  wheel,  from  center  of  track  to  center  of  track, 
is  at  all  places  one  inch  distant  from  the  track  rnade  by 


Mis(ELLAX/:()rs   I'uoni.KMs  68 

the  out?  Idc  fore  wheel.     How  many  revolutions  does  each 

wheel  of  the  buggy  make  in  going  once  round  the  track? 

Ans.  Outside  fore   wheel,  469.02+  times  ;      outside 

hind  wheel,  438.39+  times  ;     inside  front  wheel, 

406.37+  times  ;    inside  hind  wheel,  435.91+  times. 

18.  A  mechanic  converted  a  cylindrical  piece  of  iron  2 
inches  in  diameter  and  2  feet  long  into  a  wire  i  of  an  inch 
in  diameter;  what  was  the  length  of  the  Avire?    Ans.  128  ft. 

19.  Lamar  'township  purchased  tiling  for  drainage. 
Each  section  of  the  tiling  is  a  hollow  cylinder  28  inches 
high,  the  outer  diameter  being  36  inches  and  the  inner 
diameter  31  inches  ;  how  many  cubic  feet  of  material  in 
each  section,  no  allo^vance  being  made  for  the  flange? 

Ans.  4.2633+  cu.  ft. 
Note. —  The  pupil  should  work  the  above  problem  by  using  the 
area  of  the  circular  ring  as  a  base,  and  also  by  finding  the  volume 
of  two  cylinders  and  taking  their  difference, 

20.  Each  side  of  the  lower  base  of  the  frustum  of  a 
square  pyramid  is  4  feet,  of  the  upper  base  3  feet,  and 
slant -height  20  feet.  The  perimeter  of  the  lower  base 
of  the  frustum  of  a  cone  is  16  feet,  of  the  upper  base 
12  feet,  and  slant-heigat  20  feet.  How  do  the  lateral 
surfaces  compare  ?    How^  do  the  entire  surfaces  compare  ? 

Ans.  6.83+  sq.  ft.;   frus.  cone  greater. 

21.  Mr.  Johnson's  heater  has  a  cylindrical  maga- 
zine 2  feet  deep  and  10  inches  in  diameter  ;  how  many 
shovelfuls  of  coal  will  l)e  required  to  fill  it,  if  1  shovelful 
weighs  8  pounds,  and  coal  weighs  60  pounds  to  the 
cubic   foot?  Ans.  8.1812.3  shovelfuls. 

22.  Compare  the  volumes  of  a  cylinder,  a  sphere,  and 
a  cone,  whose  dimensions  are  as  follows  :  Cylindei — 
diameter  of  base  and  altitude,  each  20  inches  ;    sphere— 


64  MKNSUKATIOX 

diameter,  20  inches  ;     cone — diameter   of   base   and   alti- 
tude, each  20  inches. 

Ans.  2  vol.  cyl.=  o  vol.  sphere;     1  vol.  cyl.=  3  vol. 
cone  ;    1  vol.  sphere  =  2  vol.  cone. 

23.  Compare  the  entire  surfaces  of  the  cylinder  and 
sphere  given  in  the  preceding:  problem. 

Ans.  2  sur.  cyl.=  3  siir.  sphere. 

24.  A  globular  piece  of  wax  6  inches  in  diameter, 
a  conical  piece  whose  base  is  6  inches  and  altitude 
12  inches,  and  a  cylindrical  piece  w^hose  diameter  is 
6  inches  and  altitude  12  inches,  were  melted  and  then 
made  into  one  cubical  piece  ;  required  the  edge  of 
the    cube.  Ans.  8.26+  in. 

25.  A  farmer  placed  on  the  level  ground  a  heap  of 
apples  in  the  form  of  a  cone  whose  diameter  was  6  feet 
and  altitude  6  feet ;  this  he  covered  with  earth  of  uni- 
form thickness,  making  when  complete  a  cone  similar 
to  the  heap  of  apples  ;  the  diameter  of  the  larger  cone 
thus  formed  was  8  feet.  How  many  cubic  feet  of  earth 
in  the  covering?  Ans.  77.4928  cu.  ft. 

26.  A  granite  monument  consists  of  three  pieces  —  a 
rectangular  solid,  a  frustum  of  a  square  pyramid,  and  a 
globe  ;  the  rectangular  solid  is  6  feet  square  and  4  feet 
high  ;  the  lower  base  of  the  frustum  of  the  pyramid  is 
5  feet  square,  the  upper  base  3  feet  square,  and  altitude 
24  feet ;  the  globe  is  2  feet  in  diameter.  Required  the 
weight  of  the  entire  monument,  if  a  cubic  foot  of  granite 
weighs  168  pounds?  Ans.  45.3758+  tons. 

27.  A  hollow  squash  shaped  like  an  ellipsoid  has  an 
outside  longer  diameter  of  12  inches  and  outside  shorter 
diameter  of  8  inches  ;  the  average  thickness  of  the  sub- 
stance is  2  inches.  How  many  cubic  inches  of  material 
in   the   squash?  Ans.  335.104  cu.  in. 


.viscKLLAyKocs   i'i:(}r>ij:Ms 


(if) 


28.  The  lower  joint  of  a  stove  pipe  is  2  feet  long  ; 
the  upper  end  is  a  circle  6  inches  in  diameter,  and  the 
lower  end  an  ellipse  whose  transverse  axis  is  7  inches  and 
conjugate  axis  5  inches.  How  many  cubic  inches  of 
sand  would  be  required  to  fill  it?       Ans.  672.3024  cu.  in. 

29.  The  accompanying  diagram  represents  a  regular 
hexagon  inscribed  in  a  circle,  the  radius  of  which  is  6 
inches.  Each  side  of  a  regu- 
lar inscribed  hexagon  is  equal 
to  the  radius  of  the  circum- 
scribing circle.  Hehce,  UB  = 
6  inches,  and  any  side  of  the 
hexagon,  as  BC  =  6  inches. 

(a)    Find     the     length     of 
GH.  Ans.  5.19615+  in. 

(h)    Find  in  two  ways  the 
area  of  the  triangle  GBC. 

Ans.  15.58845  sq.  in. 

(c)  Find  in  two  ways  the  area  of  the  rhombus  ABCG. 

Ans.  31.1769  sq.  in. 

(d)  Find  in  two  ways  the  area  of  the  trapezoid  ABCD. 

Ans.  46.76535  sq.  in. 

(e)  Find  the  area  of  the  hexa- 
gon by  using  the  area  of  GBC  as 
a  basis. 

(/)  Find  the  area  of  tlie  hexa- 
gon by  using  the  area  of  ABCG  as 
a  basis. 

{(j)  Find  the  area  of  the  hexa- 
gon by  using  the  area  of  ABCD  as 
a  basis. 

(h)    Find  the  area  of  the  circumscribing  circle. 

Ans.  113.0976  sq.  in. 


Ans.  93.5307  sq.  in. 


66  M  i:\siiunox 

(l)    Fiud  the  area  of  the  6  segments. 

Ans.  19.5669  sq.  in. 

30.  A  circle  whose  radius  is  6  inches  is  circumscribed 
b}'  a  square,  and   also  has   a  square  inscribed   (Art.  48, 

Fig-.  2): 

(a)  Compare  the  length  of  the  side  of  the  circum- 
scribed square,  the  diameter  of  the  circle,  and  the 
diagonal    of    the    inscribed    square. 

(b)  Find  the  perimeter  of  the  circumscribed  square. 

Ans.  48  in. 

(c)  Find  the  circumference  of  the  circle. 

Ans.  37.6992  in. 

(d)  Find  the  perimeter  of  the  inscribed  square. 

Ans.  33.9408  in. 

(e)  Find  the  area  of  the  circumscribed  square. 

Ans.  144  sq.  in. 
(/)    Find  the  area  of  the  circle. 

Ans.  113.0976  sq.  in. 

ig)    Find  the  area  of  the  inscribed  square. 

Ans.  71.9986+  sq.  in. 

31.  In  1899  the  Williamsport  Gas  Company  erected  a 
cylindrical  vat  100  feet  in  diameter  and  100  feet  high  ; 
required  its  capacity.  Ans.  785,400  cu.  ft. 


SUPPLEMENT 

DERIVATION   OF  THE   PRISMOIDAL   FORMULA  FROM   THE 
PRINCIPLES   OF  GEOMETRY 

A  Prismoid  is  composed  of  prisms,  wedges,  and  pyra- 
mids. It  is  desired  to  find  a  common  expression  for  find- 
ing the  volume  of  these  three  volumes.  From  the  prin- 
ciples of  GeoRietry  we  get  the  following  : 


srrrfj:yi:\r 


07 


Volume  of  Prism  =  area  of  base  X  V)y  altitude. 
Volume  of  Wedgo  =  area  of  base  X  by  -^  altitude. 
Volume  of  Pyramid  =  area  of  base  X  ^^y  ^  altitude. 

area  of  either  base  of  prism, 
area  of  the  base  of  wedge, 
area  of  the  base  of  pyramid, 
length  or  altitude  of  each. 


Let  B: 

and  let  a 

Then,  Vol.  Prism  =  Ba 
Vol.  Wedge  =-^\ 
Vol.  Pyramid  =  -^* 


Rediicing  these 
three  expres- 
sions to  u 
common  de- 
nominator,we 
have  the  fol- 
lowing : 


Vol.  Prism  =  Ba  =  -(j- 
Vol.  Wedge  =  4=^- =  «.|^ 
Vol.  Pyramid  =4=*-=-!* 


Now,  let  b  = 


and  let  m 


the  area  of  a  cross -section  of  the  pyramid  at  the  apex. 

the  area  of  a  cross-section  of  the  wedge  at  its  cutting 
edge. 

the  area  of  a  cross -section  of  the  prism  at  corre- 
sponding end  of  prism. 

the  area  of  a  cross -section  of  the  pyramid  half  way 

between  base  and  apex, 
the  area  of  a  cross -section  of  the   wedge   half  way 

between  base  and  sharp  edge, 
the   area  of  a  cross -section  of  the    prism   half    way 

between  the  two  bases. 


In  the  Prism 


In  the  Wedge 


im  = 


=  B 
B 


b  =  0 


iB 


f  b  =  0 


In  the  Pyramid  -^,  ,  _. 

i  m  =  iB 

Substituting  these  values  in  the  fractions  having  a  common  denomi- 
nator, we  have  the  following  : 

Vol.Prism=«-r  =  (f-  +  t-  +  ^-r)  =  |(B-fB  +  4B)  =  ^(B-fb  +  4m). 
Vol.Wedge=T  =  (4^-  +  ^H-^-r-)=|(B-fO-f2B)=f(B+b-f4m). 
Vol. Pyramid  =^l:^=(4^-+^-f  4^-)=  ^(B-fO  +  B  )  =  f  (B+ b  +  4m). 
Hence,  the  volume  of  the  Prism,  or  Wedge,  or  Pyramid  = 
§  (B -|- b -|- 4m),  and  the  volume  of  a  Prismoid,  therefore  = 
|(B4-b-f-4m). 


(l.s  MJty.suiiATioy 

The  rule  may  be  expressed  in  words  as  follows  :  M^d' 
tiply  the  sum  of  both  end  areas  and  four  times  the  area 
of  a  section  half  way  between  them,  by  one-sixth  the  altitude. 

A  cylinder  may  be  divided  into  an  infinite  number  of 
prisms  ;  a  sphere  or  a  cone  into  an  infinite  number  of 
pyramids  ;  frustums  of  pyramids  and  cones  into  prisms 
and  p3^ramids,  etc.  It  is  evident,  therefore,  that  this  rule 
is  very  wide  in  its  application. 

If,  for  the  purpose  of  association,  w^e  consider  the 
capital  letter  B  as  the  initial  letter  of  the  term  "Base" 
(large),  and  the  small  letter  b  as  the  initial  letter  of 
"base"  (small),  and  m  the  initial  letter  of  the  term 
"middle  section,"  the  rule  may  be  quite  easily  remem- 
bered. In  some  solids,  as  the  pyramid,  cone,  and  their 
frustums,  these  letters  will  have  special  significance,  since 
one  base  is  larger  than  the  other.  This  is  not  true,  how- 
ever, of  all  solids. 

The  following  abbreviated  process  of  using  the  formula 
may  be  used  to  advantage  after  the  pupil  has  become 
thoroughly  grounded  in  the  process  used  in  the  chapter 
on   Illustrative  Problems. 

Problem.— What  are  the  solid  contents  of  the  frustum 

of  a  square  pyramid  whose  upper  base  is  6   feet  square, 

lower  base  10  feet  square,  and  altitude  30  feet  f 

Operation. — 

B=  lOX  10  =100  sq.  ft. 
b=        GX    6  =    36 

4m  =  4  (8X    8)  =  256 

392  X¥  =  1,960    eu.  ft.,  Ans. 

Probleim.— What    is    the   volume   of   a   sphere   whose 

radius  is  10  feet"? 

Operation. — 

B=  0.0 

b=  0.0 

4iii  =  47rr-=  4(3.1416  X 100)  =1,256.64 

1, 256. 54X¥  =  4,188.8  eu.  ft.,  Ans. 


MEASURES    OF   EXTENSION 


69 


MEASURES  OF  EXTENSION 

LINEAR  MEASURE 

TABLE 

12     inches  (in.)  =  1  foot  (ft.) 
3     feet  =  1  yard  (yd.) 

5i  yards 

16i  feet 
320     rods  =1  mile  (mi.) 


}  = 


1  rod  (rd.) 


1  mi.=  320  rds.=  1,700  yds.=  5,280  ft. =  63,360  in. 
Scale.— 320,  5^,  3,  12. 


OTHER    DENOMINATIONS 


3  barleycorns 

4  inches 

9  inches 

8  spans 

6  feet 

18  inches 

21.888  inches 

20  fathoms 

3  feet 

8  fnrlongs 

1.152f  common  miles 

3  geographical  miles 
60  geographical,  or  | 
69.16  statute  miles  j 
360  d'"2rees 


==  1   inch   (Used  by  shoemakers) 

=  1   hand    (Used  to  measure  the 
height  of    horses) 

=  1  span 

=  1  fathom     (Used  by  sailors) 

=  1   fathom   (Used  to  measure  the 
depths  at  sea) 

=  1  common  cubit 

=  1  sacred  cubit 

=  1  cable's  length 

=  1  pace 

=  1  mile 

=  1  geographical,  or  nautical 

mile  =  1  knot 
=  1  league 

r  of  latitude  on  a  me- 

=  1    degree  "^  ridian,  or  of  longi- 

(_  tude  on  the  equatoi 

=  the    circumference    of    the 

earth 


70  MENS  URA  TION 

SURVEYORS'  LINEAR   MEASURE 

TABLE 

7.92  inches  =1  link  (1.) 
25  links  =1  rod  (rd.) 

4  rods  =1  chain  (ch.) 

80  chains       =1  mile  (mi.) 
1  mi.=  80  eh.=  320  rds.=  8,000  1.=  63,360  in. 

A  Gunter's  Chain  is  the  unit  of  measure,  and  is  4  rds.  or  66  ft. 
long,  and  consists  of  100  links. 

SQUARE  MEASURE 

TABLE 

144    square  inches  (sq.  in.)   =1  square  foot  (sq.  ft.) 
9    square  feet  =  1  square  yard  (sq.  yd.) 

3O4  square  yards  =  1  square  rod  or  perch 

160    square  rods  or  perches=  1  acre  (A.)  [(sq.  rd.,P.) 

640    acres  =  1  square  mile  (sq.  mi.) 

1  sq.  mi.=  640  A.=  102,400  sq.  rds.=  3,097,600  sq.  yds.=  - 
27,878,400  sq.  ft. =  4,014,489,600  sq.  in. 

Scale.— 640,  160,  30i,  9,  144. 

SURVEYORS'  SQUARE  MEASURE 

TABLE 

625  square  links  (sq.  1.)  =1  pole  (P.) 

1(3  poles  =1  square  chain  (sq.ch.) 

10  square  chains  =1  acre  (A.) 

640  acres  =  1  square  mile,  or  sec- 

tion  (sq.  mi.,  see.) 
36  square  miles  (6  miles  square)  =  1  township  (Tp.) 

1  Tp.=  36  sq.  mi. =  23,040  A. =  230,400  sq.  eh. =  3,686,400  P.= 
2,304,000,000  sq.  1. 

Scale.—  36,  640,  10,  16,  625. 


MI':ASiHES.OF    fJXTWSfO^  ,  71 


CUBIC    MEASURE 

TABLE 


1,728     cubic  inches  (cu.  in.)  =  l  cubic  foot  (cu.  ft.) 

27     cubic  feet  =  1  cubic  j'ard  (cu.  yd.) 

10     cubic  feet  =  1  cord  foot  (cd.  ft.) 

8     cord  feet,  or)  ^  ,     „  i   ,      . 

y  =1  cord  of  wood  (cd.) 


128     cubic  feet 


24i  cubic  feet  =  1  {  P^''^^  °*  ^^°"«'  |  (Pch.) 

I  or  or  masonry  ) 


1  cu.  yd.=  27  cu.  ft.=  46,656  cu.  in. 
Scale.— 27,  1,728. 


CIRCULAR  OR  AKGULAR  MEASURE 

TABLE 

60  seconds  (")  =1  minute  (') 

60  minutes  =  1  degree  (°) 

30  degrees  =1  sign  (S.) 

12  signs,  or  360  degrees  =  1  circle  (cir.) 

1  eir.=  12  S.=  30°=  21,600=  1,296,000". 
Scale.— 12,  30,  60,  60. 

DUODECIMALS 

TABLE 

12  fourths  ("")  =  1  third  ('") 

12  thirds  =1  second  ( ') 

12  seconds  =  1  prime  (') 

12  primes,  or  inches  =  1  foot  (ft.) 

1  ft.=  12  '=  144"=  1 ,728'"==  20,736"". 
Scale.—  12,  12,  12,  12. 


72  MEXSUR.ITION 

MISCELLANEOUS 
2,150.42  cubic  inches  =  1  Winchester  bushel  (bu.) 

COMPARATIVE   TABLE   OF  WEIGHTS 

Atwirdupois  Troy  Apothecaries' 

1  pound  =  7,000      grains  =  5,760  grains  =  5,760  grains 
1  ounce  =     437.5  grains  =     480  grains  =     480  grains 

COMPARATIVE   TABLE   OF  MEASURES   OF  CAPACITY 

C'u.  in.  in  Cu.  in.  in  Cu.  in.  in  Cu.  in.  in 

one  gallon         one  quart  one  pint  one  gill 

Liquid  Measure 231  574  288^  lit^ 

Dry  Measure  (half -peck)  .    2685  675  33  5^  St 


UNIVERSITY   OF   CALIFORNIA   L 


IBRARY 


THIS  BOOK  IS  DUE  ON  THE  LAST  DATE 
STAMPED  BELOW 


^<AH^4.W18 


l-i-*?© 


NOV    2Q]92i^ 


otc 


4    ^9G8 


RECEIVED 


LOAN  DEPtJ 

NOV  111976    * 


WAR    4)986 

**     WR2  4B86 


30»t-6,'14 


V^^^^       ^  0 


'"'^'^'^^UBmy.u.c., ,„,,,,, 


260008 


